30 Stoichiometry Problems to Practice On

How do you actually get good at stoichiometry? Not by reading about it. By doing it, over and over, until the pattern clicks.

Balance the equation. Find the mole ratio. Convert, convert, convert. That's the whole game, and it becomes automatic once you've run it enough times.

So that's what we're doing here: thirty problems, starting with simple mole-ratio questions and building up to limiting reactant and percent yield.

One rule. Try each problem yourself before you open the solution. That's where the real learning happens, not in reading someone else's steps. Every problem below has a Show step-by-step solution toggle with the full reasoning and the final answer.

Section 1: Mole-to-Mole Conversions

Start here. You're given moles of one substance and asked for moles of another, and the coefficients in the balanced equation are the only tool you need. Get comfortable with this move first. Everything below just adds a conversion step or two on top of it.

Problem 1

Given the reaction $N_2 + 3H_2 \rightarrow 2NH_3$ , how many moles of $NH_3$ are produced from 4 moles of $N_2$ ?

Step 1: Identify the mole ratio. From the balanced equation, the coefficient ratio of $N_2$ to $NH_3$ is $1:2$ .

Step 2: Apply the ratio. $4 \text{ mol } N_2 \times \frac{2 \text{ mol } NH_3}{1 \text{ mol } N_2} = 8 \text{ mol } NH_3$

Answer8 mol NH₃

Problem 2

For $2H_2 + O_2 \rightarrow 2H_2O$ , how many moles of $O_2$ are needed to react with 6 moles of $H_2$ ?

Step 1: Identify the mole ratio. $H_2 : O_2 = 2:1$ .

Step 2: Apply the ratio. $6 \text{ mol } H_2 \times \frac{1 \text{ mol } O_2}{2 \text{ mol } H_2} = 3 \text{ mol } O_2$

Answer3 mol O₂

Problem 3

Given $2Al + 3Cl_2 \rightarrow 2AlCl_3$ , how many moles of $AlCl_3$ form from 5 moles of $Al$ ?

Step 1: Identify the mole ratio. $Al : AlCl_3 = 2:2 = 1:1$ .

Step 2: Apply the ratio. $5 \text{ mol } Al \times \frac{2 \text{ mol } AlCl_3}{2 \text{ mol } Al} = 5 \text{ mol } AlCl_3$

Answer5 mol AlCl₃

Problem 4

For $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ , how many moles of $CO_2$ are produced from 2 moles of $C_3H_8$ ?

Step 1: Identify the mole ratio. $C_3H_8 : CO_2 = 1:3$ .

Step 2: Apply the ratio. $2 \text{ mol } C_3H_8 \times \frac{3 \text{ mol } CO_2}{1 \text{ mol } C_3H_8} = 6 \text{ mol } CO_2$

Answer6 mol CO₂

Problem 5

Given $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ , how many moles of $Fe$ are needed to produce 8 moles of $Fe_2O_3$ ?

Step 1: Identify the mole ratio. $Fe : Fe_2O_3 = 4:2 = 2:1$ .

Step 2: Apply the ratio. $8 \text{ mol } Fe_2O_3 \times \frac{4 \text{ mol } Fe}{2 \text{ mol } Fe_2O_3} = 16 \text{ mol } Fe$

Answer16 mol Fe

Problem 6

For $2KClO_3 \rightarrow 2KCl + 3O_2$ , how many moles of $O_2$ form from 10 moles of $KClO_3$ ?

Step 1: Identify the mole ratio. $KClO_3 : O_2 = 2:3$ .

Step 2: Apply the ratio. $10 \text{ mol } KClO_3 \times \frac{3 \text{ mol } O_2}{2 \text{ mol } KClO_3} = 15 \text{ mol } O_2$

Answer15 mol O₂

Section 2: Mole-to-Mass and Mass-to-Mole Conversions

Now we bring molar mass into it. You'll convert grams to moles, or moles to grams, using the periodic table. Then it's the exact same ratio logic from Section 1. New step, not a new idea.

Problem 7

How many grams of $NH_3$ are produced from 3 moles of $N_2$ in $N_2 + 3H_2 \rightarrow 2NH_3$ ?

Step 1: Mole ratio. $N_2 : NH_3 = 1:2$ , so 3 mol $N_2$ gives 6 mol $NH_3$ .

Step 2: Molar mass of $NH_3$ . $14.01 + 3(1.01) = 17.03$ g/mol.

Step 3: Convert to grams. $6 \text{ mol } NH_3 \times 17.03 \text{ g/mol} = 102.2 \text{ g}$

Answer≈102 g NH₃

Problem 8

How many moles of $O_2$ are needed to completely react with 50 g of $H_2$ in $2H_2 + O_2 \rightarrow 2H_2O$ ?

Step 1: Convert grams of $H_2$ to moles. Molar mass of $H_2$ = 2.016 g/mol. $50 \text{ g} \div 2.016 \text{ g/mol} = 24.80 \text{ mol } H_2$

Step 2: Apply the mole ratio. $H_2 : O_2 = 2:1$ . $24.80 \text{ mol } H_2 \times \frac{1 \text{ mol } O_2}{2 \text{ mol } H_2} = 12.4 \text{ mol } O_2$

Answer≈12.4 mol O₂

Problem 9

Given $2Mg + O_2 \rightarrow 2MgO$ , how many grams of $MgO$ form from 4 moles of $Mg$ ?

Step 1: Mole ratio. $Mg : MgO = 1:1$ , so 4 mol $Mg$ gives 4 mol $MgO$ .

Step 2: Molar mass of $MgO$ . $24.31 + 16.00 = 40.31$ g/mol.

Step 3: Convert to grams. $4 \text{ mol } MgO \times 40.31 \text{ g/mol} = 161.2 \text{ g}$

Answer≈161 g MgO

Problem 10

How many moles of $CO_2$ are produced from 88 g of $C_3H_8$ burning completely in $C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O$ ?

Step 1: Convert grams of $C_3H_8$ to moles. Molar mass = $3(12.01) + 8(1.01) = 44.1$ g/mol. $88 \text{ g} \div 44.1 \text{ g/mol} = 2.00 \text{ mol } C_3H_8$

Step 2: Apply the mole ratio. $C_3H_8 : CO_2 = 1:3$ . $2.00 \text{ mol} \times 3 = 6.00 \text{ mol } CO_2$

Answer6 mol CO₂

Problem 11

For $2Na + Cl_2 \rightarrow 2NaCl$ , how many grams of $NaCl$ form from 0.75 moles of $Na$ ?

Step 1: Mole ratio. $Na : NaCl = 2:2 = 1:1$ , so 0.75 mol $Na$ gives 0.75 mol $NaCl$ .

Step 2: Molar mass of $NaCl$ . $22.99 + 35.45 = 58.44$ g/mol.

Step 3: Convert to grams. $0.75 \text{ mol} \times 58.44 \text{ g/mol} = 43.8 \text{ g}$

Answer≈43.8 g NaCl

Problem 12

Given $4Al + 3O_2 \rightarrow 2Al_2O_3$ , how many moles of $Al$ are needed to produce 51 g of $Al_2O_3$ ?

Step 1: Convert grams of $Al_2O_3$ to moles. Molar mass = $2(26.98) + 3(16.00) = 101.96$ g/mol. $51 \text{ g} \div 101.96 \text{ g/mol} = 0.50 \text{ mol } Al_2O_3$

Step 2: Apply the mole ratio. $Al : Al_2O_3 = 4:2 = 2:1$ . $0.50 \text{ mol} \times 2 = 1.0 \text{ mol } Al$

Answer1.0 mol Al

Problem 13

Given $CaCO_3 \rightarrow CaO + CO_2$ , how many grams of $CO_2$ form from 2.5 moles of $CaCO_3$ ?

Step 1: Mole ratio. $CaCO_3 : CO_2 = 1:1$ , so 2.5 mol $CaCO_3$ gives 2.5 mol $CO_2$ .

Step 2: Molar mass of $CO_2$ . $12.01 + 2(16.00) = 44.01$ g/mol.

Step 3: Convert to grams. $2.5 \text{ mol} \times 44.01 \text{ g/mol} = 110.0 \text{ g}$

Answer110 g CO₂

Problem 14

Given $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ , how many moles of $H_2SO_4$ are needed to fully react with 40 g of $NaOH$ ?

Step 1: Convert grams of $NaOH$ to moles. Molar mass = $22.99 + 16.00 + 1.01 = 40.00$ g/mol. $40 \text{ g} \div 40.00 \text{ g/mol} = 1.0 \text{ mol } NaOH$

Step 2: Apply the mole ratio. $H_2SO_4 : NaOH = 1:2$ . $1.0 \text{ mol } NaOH \times \frac{1 \text{ mol } H_2SO_4}{2 \text{ mol } NaOH} = 0.5 \text{ mol } H_2SO_4$

Answer0.5 mol H₂SO₄

Section 3: Mass-to-Mass Conversions

This is the version you'll actually use outside a classroom. Grams in, grams out. Which means three steps instead of one or two: convert to moles, apply the ratio, convert back.

Problem 15

How many grams of $H_2O$ form from 20 g of $H_2$ reacting completely with $O_2$ in $2H_2 + O_2 \rightarrow 2H_2O$ ?

Step 1: Convert grams of $H_2$ to moles. Molar mass = 2.016 g/mol. $20 \text{ g} \div 2.016 \text{ g/mol} = 9.92 \text{ mol } H_2$

Step 2: Apply the mole ratio. $H_2 : H_2O = 2:2 = 1:1$ . $9.92 \text{ mol } H_2 = 9.92 \text{ mol } H_2O$

Step 3: Convert moles of $H_2O$ to grams. Molar mass = 18.02 g/mol. $9.92 \text{ mol} \times 18.02 \text{ g/mol} = 178.8 \text{ g}$

Answer≈178.8 g H₂O

Problem 16

Given $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$ , how many grams of $CO_2$ form from 30 g of $C_2H_6$ ?

Step 1: Convert grams of $C_2H_6$ to moles. Molar mass = $2(12.01) + 6(1.01) = 30.07$ g/mol. $30 \text{ g} \div 30.07 \text{ g/mol} = 0.998 \text{ mol } C_2H_6$

Step 2: Apply the mole ratio. $C_2H_6 : CO_2 = 2:4 = 1:2$ . $0.998 \text{ mol} \times 2 = 1.996 \text{ mol } CO_2$

Step 3: Convert moles of $CO_2$ to grams. Molar mass = 44.01 g/mol. $1.996 \text{ mol} \times 44.01 \text{ g/mol} = 87.8 \text{ g}$

Answer≈87.8 g CO₂

Problem 17

Given $4Fe + 3O_2 \rightarrow 2Fe_2O_3$ , how many grams of $Fe_2O_3$ form from 100 g of $Fe$ ?

Step 1: Convert grams of $Fe$ to moles. Molar mass = 55.85 g/mol. $100 \text{ g} \div 55.85 \text{ g/mol} = 1.791 \text{ mol } Fe$

Step 2: Apply the mole ratio. $Fe : Fe_2O_3 = 4:2 = 2:1$ . $1.791 \text{ mol } Fe \times \frac{1 \text{ mol } Fe_2O_3}{2 \text{ mol } Fe} = 0.896 \text{ mol } Fe_2O_3$

Step 3: Convert moles of $Fe_2O_3$ to grams. Molar mass = $2(55.85) + 3(16.00) = 159.70$ g/mol. $0.896 \text{ mol} \times 159.70 \text{ g/mol} = 143.1 \text{ g}$

Answer≈143 g Fe₂O₃

Problem 18

For $N_2 + 3H_2 \rightarrow 2NH_3$ , how many grams of $NH_3$ form from 28 g of $N_2$ ?

Step 1: Convert grams of $N_2$ to moles. Molar mass = 28.02 g/mol. $28 \text{ g} \div 28.02 \text{ g/mol} = 1.00 \text{ mol } N_2$

Step 2: Apply the mole ratio. $N_2 : NH_3 = 1:2$ . $1.00 \text{ mol} \times 2 = 2.00 \text{ mol } NH_3$

Step 3: Convert moles of $NH_3$ to grams. Molar mass = 17.03 g/mol. $2.00 \text{ mol} \times 17.03 \text{ g/mol} = 34.1 \text{ g}$

Answer≈34.1 g NH₃

Problem 19

Given $2AgNO_3 + CaCl_2 \rightarrow 2AgCl + Ca(NO_3)_2$ , how many grams of $AgCl$ precipitate form from 34 g of $AgNO_3$ ?

Step 1: Convert grams of $AgNO_3$ to moles. Molar mass = $107.87 + 14.01 + 3(16.00) = 169.87$ g/mol. $34 \text{ g} \div 169.87 \text{ g/mol} = 0.2001 \text{ mol } AgNO_3$

Step 2: Apply the mole ratio. $AgNO_3 : AgCl = 2:2 = 1:1$ . $0.2001 \text{ mol } AgNO_3 = 0.2001 \text{ mol } AgCl$

Step 3: Convert moles of $AgCl$ to grams. Molar mass = $107.87 + 35.45 = 143.32$ g/mol. $0.2001 \text{ mol} \times 143.32 \text{ g/mol} = 28.7 \text{ g}$

Answer≈28.7 g AgCl

Problem 20

How many grams of $CO_2$ are released when 500 g of $CaCO_3$ fully decomposes in $CaCO_3 \rightarrow CaO + CO_2$ ?

Step 1: Convert grams of $CaCO_3$ to moles. Molar mass = $40.08 + 12.01 + 3(16.00) = 100.09$ g/mol. $500 \text{ g} \div 100.09 \text{ g/mol} = 5.00 \text{ mol } CaCO_3$

Step 2: Apply the mole ratio. $CaCO_3 : CO_2 = 1:1$ , so 5.00 mol $CaCO_3$ gives 5.00 mol $CO_2$ .

Step 3: Convert moles of $CO_2$ to grams. Molar mass = 44.01 g/mol. $5.00 \text{ mol} \times 44.01 \text{ g/mol} = 220.1 \text{ g}$

Answer≈220 g CO₂

Section 4: Limiting Reactant Problems

What happens when a reaction has two reactants instead of one? They almost never run out at the same time. One of them hits zero first, and that one, the limiting reactant, is the only one that matters. It controls how much product you get, no matter how much of the other reactant is left sitting around.

Problem 21

Given $N_2 + 3H_2 \rightarrow 2NH_3$ , if you start with 4 moles of $N_2$ and 9 moles of $H_2$ , which reactant is limiting?

Step 1: Find how much $H_2$ the available $N_2$ would need. Ratio $N_2 : H_2 = 1:3$ . $4 \text{ mol } N_2 \times 3 = 12 \text{ mol } H_2 \text{ required}$

Step 2: Compare to what's available. Only 9 mol $H_2$ is available, but 12 mol is needed. $H_2$ runs out first.

AnswerH₂ is the limiting reactant.

Problem 22

For $2H_2 + O_2 \rightarrow 2H_2O$ , if you have 10 g of $H_2$ and 10 g of $O_2$ , which is the limiting reactant, and how many grams of $H_2O$ form?

Step 1: Convert both reactants to moles. $10 \text{ g } H_2 \div 2.016 \text{ g/mol} = 4.96 \text{ mol } H_2$ $10 \text{ g } O_2 \div 32.00 \text{ g/mol} = 0.3125 \text{ mol } O_2$

Step 2: Find how much $O_2$ the available $H_2$ would need. Ratio $H_2 : O_2 = 2:1$ . $4.96 \text{ mol } H_2 \times \frac{1}{2} = 2.48 \text{ mol } O_2 \text{ required}$

Step 3: Compare to what's available. Only 0.3125 mol $O_2$ is available, far short of the 2.48 mol needed. $O_2$ is limiting.

Step 4: Calculate $H_2O$ from the limiting reactant. Ratio $O_2 : H_2O = 1:2$ . $0.3125 \text{ mol } O_2 \times 2 = 0.625 \text{ mol } H_2O$ $0.625 \text{ mol} \times 18.02 \text{ g/mol} = 11.3 \text{ g}$

AnswerO₂ is limiting; ≈11.3 g H₂O forms

Problem 23

Given $2Al + 3Cl_2 \rightarrow 2AlCl_3$ , if you react 5.4 g of $Al$ with 20 g of $Cl_2$ , which is limiting, and how many grams of $AlCl_3$ form?

Step 1: Convert both reactants to moles. $5.4 \text{ g } Al \div 26.98 \text{ g/mol} = 0.200 \text{ mol } Al$ $20 \text{ g } Cl_2 \div 70.90 \text{ g/mol} = 0.282 \text{ mol } Cl_2$

Step 2: Find how much $Cl_2$ the available $Al$ would need. Ratio $Al : Cl_2 = 2:3$ . $0.200 \text{ mol } Al \times \frac{3}{2} = 0.300 \text{ mol } Cl_2 \text{ required}$

Step 3: Compare to what's available. Only 0.282 mol $Cl_2$ is available, but 0.300 mol is needed. $Cl_2$ runs out first and is limiting, not $Al$ , even though $Cl_2$ 's mass looks bigger here. Always check by moles and ratio, not by which mass looks bigger.

Step 4: Calculate $AlCl_3$ from the limiting reactant. Ratio $Cl_2 : AlCl_3 = 3:2$ . $0.282 \text{ mol } Cl_2 \times \frac{2}{3} = 0.188 \text{ mol } AlCl_3$ $0.188 \text{ mol} \times 133.33 \text{ g/mol} = 25.1 \text{ g}$

AnswerCl₂ is limiting; ≈25.1 g AlCl₃ forms

Problem 24

For $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ , if you burn 16 g of $CH_4$ with 48 g of $O_2$ , is there excess reactant left over, and if so, how much?

Step 1: Convert both reactants to moles. $16 \text{ g } CH_4 \div 16.04 \text{ g/mol} = 0.998 \text{ mol } CH_4$ $48 \text{ g } O_2 \div 32.00 \text{ g/mol} = 1.50 \text{ mol } O_2$

Step 2: Find how much $O_2$ the available $CH_4$ would need. Ratio $CH_4 : O_2 = 1:2$ . $0.998 \text{ mol } CH_4 \times 2 = 2.00 \text{ mol } O_2 \text{ required}$

Step 3: Compare to what's available. Only 1.50 mol $O_2$ is available, but 2.00 mol is needed. $O_2$ is limiting, which means $CH_4$ is the one left over in excess.

Step 4: Find how much $CH_4$ actually reacts with the available $O_2$ . Ratio $O_2 : CH_4 = 2:1$ . $1.50 \text{ mol } O_2 \times \frac{1}{2} = 0.75 \text{ mol } CH_4 \text{ reacted}$

Step 5: Subtract to find the leftover $CH_4$ . $0.998 \text{ mol} - 0.75 \text{ mol} = 0.248 \text{ mol } CH_4 \text{ remaining}$ $0.248 \text{ mol} \times 16.04 \text{ g/mol} = 4.0 \text{ g}$

AnswerO₂ is limiting; ≈4.0 g of CH₄ is left over unreacted

Problem 25

Given $2Na + Cl_2 \rightarrow 2NaCl$ , if 46 g of $Na$ reacts with 71 g of $Cl_2$ , how many grams of $NaCl$ are produced?

Step 1: Convert both reactants to moles. $46 \text{ g } Na \div 22.99 \text{ g/mol} = 2.00 \text{ mol } Na$ $71 \text{ g } Cl_2 \div 70.90 \text{ g/mol} = 1.00 \text{ mol } Cl_2$

Step 2: Check the ratio needed. $Na : Cl_2 = 2:1$ , and $2.00 : 1.00$ is exactly that ratio, so both reactants run out at the same time. This is the rare case where there's no true "excess" reactant, so the product can be calculated from either one.

Step 3: Calculate $NaCl$ . Ratio $Na : NaCl = 2:2 = 1:1$ . $2.00 \text{ mol } Na \times 1 = 2.00 \text{ mol } NaCl$ $2.00 \text{ mol} \times 58.44 \text{ g/mol} = 116.9 \text{ g}$

Answer≈117 g NaCl

Problem 26

For $Zn + 2HCl \rightarrow ZnCl_2 + H_2$ , if 13 g of $Zn$ reacts with 20 g of $HCl$ , how many grams of $H_2$ gas form?

Step 1: Convert both reactants to moles. $13 \text{ g } Zn \div 65.38 \text{ g/mol} = 0.199 \text{ mol } Zn$ $20 \text{ g } HCl \div 36.46 \text{ g/mol} = 0.548 \text{ mol } HCl$

Step 2: Find how much $HCl$ the available $Zn$ would need. Ratio $Zn : HCl = 1:2$ . $0.199 \text{ mol } Zn \times 2 = 0.398 \text{ mol } HCl \text{ required}$

Step 3: Compare to what's available. 0.548 mol $HCl$ is available, more than the 0.398 mol needed. $HCl$ is in excess, so $Zn$ is limiting.

Step 4: Calculate $H_2$ from the limiting reactant. Ratio $Zn : H_2 = 1:1$ . $0.199 \text{ mol } Zn = 0.199 \text{ mol } H_2$ $0.199 \text{ mol} \times 2.016 \text{ g/mol} = 0.40 \text{ g}$

AnswerZn is limiting; ≈0.40 g H₂ forms

Section 5: Percent Yield Problems

Here's something the balanced equation won't tell you: real reactions never produce quite as much product as the math predicts. Percent yield measures exactly how much less. It's the gap between what the equation promises (the theoretical yield) and what you actually collect (the actual yield), written as a percentage.

Problem 27

For $2H_2 + O_2 \rightarrow 2H_2O$ , the theoretical yield of water from a reaction is 36 g, but only 30 g is actually collected. What is the percent yield?

Step 1: Recall the percent yield formula. $\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100$

Step 2: Plug in the values. $\frac{30 \text{ g}}{36 \text{ g}} \times 100 = 83.3\%$

Answer83.3%

Problem 28

Given $N_2 + 3H_2 \rightarrow 2NH_3$ , 5 moles of $N_2$ reacts completely with excess $H_2$ . If only 6.8 moles of $NH_3$ are collected, what is the percent yield?

Step 1: Calculate the theoretical yield of $NH_3$ . Ratio $N_2 : NH_3 = 1:2$ . $5 \text{ mol } N_2 \times 2 = 10 \text{ mol } NH_3 \text{ (theoretical)}$

Step 2: Apply the percent yield formula. $\frac{6.8 \text{ mol}}{10 \text{ mol}} \times 100 = 68\%$

Answer68%

Problem 29

For $CaCO_3 \rightarrow CaO + CO_2$ , 250 g of $CaCO_3$ decomposes and produces 96 g of $CO_2$ . What is the percent yield?

Step 1: Convert grams of $CaCO_3$ to moles. Molar mass = 100.09 g/mol. $250 \text{ g} \div 100.09 \text{ g/mol} = 2.498 \text{ mol } CaCO_3$

Step 2: Calculate theoretical moles and mass of $CO_2$ . Ratio $CaCO_3 : CO_2 = 1:1$ . $2.498 \text{ mol } CO_2 \times 44.01 \text{ g/mol} = 109.9 \text{ g (theoretical)}$

Step 3: Apply the percent yield formula. $\frac{96 \text{ g}}{109.9 \text{ g}} \times 100 = 87.3\%$

Answer≈87.3%

Problem 30

Given $2C_2H_6 + 7O_2 \rightarrow 4CO_2 + 6H_2O$ , 60 g of $C_2H_6$ reacts completely with excess $O_2$ . If the actual yield of $CO_2$ is 150 g, what is the percent yield?

Step 1: Convert grams of $C_2H_6$ to moles. Molar mass = 30.07 g/mol. $60 \text{ g} \div 30.07 \text{ g/mol} = 1.996 \text{ mol } C_2H_6$

Step 2: Calculate theoretical moles and mass of $CO_2$ . Ratio $C_2H_6 : CO_2 = 2:4 = 1:2$ . $1.996 \text{ mol} \times 2 = 3.992 \text{ mol } CO_2$ $3.992 \text{ mol} \times 44.01 \text{ g/mol} = 175.7 \text{ g (theoretical)}$

Step 3: Apply the percent yield formula. $\frac{150 \text{ g}}{175.7 \text{ g}} \times 100 = 85.4\%$

Answer≈85.4%

Molar masses used throughout are rounded to two decimal places from standard atomic weights. If your textbook or instructor uses slightly different rounding, your answers may differ by a few tenths of a percent. That's expected, and not a sign of a mistake in your method.

30 Stoichiometry Problems to Practice On (With Step-by-Step Solutions)