Stoichiometry: practice problems with step-by-step solutions
Free Stoichiometry practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: mass to mass, limiting reactant, percent yield.
Free practice problemEasy
For the reaction C3H8 + 5O2 → 3CO2 + 4H2O, what mass m of H2O forms from 100 g of C3H8? Assume the other reactants are in excess.
Use the coefficients as the mole ratio to find the moles of the substance you want.
Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.
Common mistakes
Using the mass ratio instead of the mole ratio.
Forgetting to balance the equation.
Not checking which reactant is limiting.
Key formulas
Moles
n = m/M
Concentration
c = n/V
Mole ratio
use the coefficients of the balanced equation
Limiting reactant
the one that runs out first
Percent yield
actual / theoretical × 100%
Learn Stoichiometry from scratch
1What stoichiometry is
A chemical equation is a recipe. 2H₂ + O₂ → 2H₂O says that two molecules of hydrogen react with one molecule of oxygen to give two molecules of water.
Stoichiometry uses that recipe to answer “how much?” questions: how much product you get from a given amount of starting material, or how much you need.
The numbers in front of the formulas are the coefficients. They count particles, not grams. The equation must be balanced: every atom appears the same number of times on both sides.
Particles are far too small to count one by one, so chemists count them in moles. One mole is 6.022 × 10²³ particles, just as one dozen is 12 eggs.
So the coefficients also give a mole ratio: 2 mol of H₂ reacts with 1 mol of O₂ and gives 2 mol of H₂O.
Self-check In N₂ + 3H₂ → 2NH₃, how many moles of NH₃ form from 2 mol of N₂ (with enough H₂)?
1 mol
2 mol
4 mol
6 mol
Answer: 4 mol. 1 mol of N₂ gives 2 mol of NH₃, so 2 mol of N₂ gives 4 mol. Giving 2 mol ignores the mole ratio, and 6 mol uses the coefficient of H₂ instead.
2Moles and molar mass
A balance measures grams, but the equation works in moles. The molar mass M links the two: it is the mass of one mole, in g/mol.
n = m / M m = n · M
You find M by adding the atomic masses from the periodic table. With H = 1.0, C = 12.0 and O = 16.0:
M(H₂O) = 2 · 1.0 + 16.0 = 18.0 g/mol
M(CO₂) = 12.0 + 2 · 16.0 = 44.0 g/mol
M(CH₄) = 12.0 + 4 · 1.0 = 16.0 g/mol
Example: 36 g of water is 36 / 18.0 = 2.0 mol of water.
Exam questions usually give the molar masses, with two decimals. Use the values given.
Self-check How many moles are there in 88 g of CO₂ (M = 44 g/mol)?
0.5 mol
2 mol
3872 mol
88 mol
Answer: 2 mol. n = m/M = 88/44 = 2 mol. Dividing the other way (44/88 = 0.5) or multiplying (88 · 44) uses the formula upside down.
3The method step by step
Every mass-to-mass question follows the same road: from grams to moles, across the equation with the mole ratio, and back to grams.
Check that the equation is balanced.
Convert the given mass of A to moles: nA = mA / MA.
Use the coefficients: nB = nA · b / a, where a and b are the coefficients of A and B.
Convert back to mass: mB = nB · MB.
With two reactants given, first find the limiting reactant and start from that one.
For percent yield, compare the mass actually collected with the mass you calculated.
4A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
For the reaction CH₄ + 2O₂ → CO₂ + 2H₂O, what mass m of H₂O forms from 8.0 g of CH₄? Assume the oxygen is in excess. Use M(CH₄) = 16.0 g/mol and M(H₂O) = 18.0 g/mol.
1
Check the equation
Left: 1 C, 4 H, 4 O. Right: 1 C, 4 H and 2 + 2 = 4 O. The equation is balanced.
2
Mass to moles
n(CH₄) = 8.0 / 16.0 = 0.50 mol
3
Use the mole ratio
1 mol of CH₄ gives 2 mol of H₂O, so the factor is 2/1.
n(H₂O) = 0.50 · 2 / 1 = 1.0 mol
4
Moles to mass
m(H₂O) = 1.0 · 18.0 = 18 g
5
Check the answer
8 g of CH₄ needs 1.0 mol of O₂, which is 32 g. Then 8 + 32 = 40 g goes in. 0.50 mol of CO₂ is 22 g, and 22 + 18 = 40 g comes out. Mass is conserved, so the answer fits.
Answer
m(H₂O) = 18 g
5Limiting reactant and percent yield
When two reactants are given, one of them usually runs out first. That is the limiting reactant. It decides how much product forms; the other is left over (in excess).
To find it, divide the moles of each reactant by its coefficient. The smaller result belongs to the limiting reactant.
Example: 4.0 g of H₂ (M = 2.0) and 16 g of O₂ (M = 32) react in 2H₂ + O₂ → 2H₂O. That is 2.0 mol of H₂ and 0.50 mol of O₂.
So 0.50 mol of O₂ gives 0.50 · 2 = 1.0 mol of water, which is 18 g. Part of the hydrogen is left over.
In practice you never collect all the product. The percent yield compares what you got with what you calculated, the theoretical yield:
Y = actual / theoretical × 100%
Self-check The theoretical yield of a reaction is 18 g, and you collect 15.3 g. What is the percent yield?
118%
15.3%
2.7%
85%
Answer: 85%. Y = 15.3/18 × 100% = 85%. Dividing the other way gives 118%, which is impossible: you cannot collect more than the theoretical amount.
6Common mistakes
Using the mass ratio instead of the mole ratio. The coefficients count moles, never grams.
Working with an equation that is not balanced.
Starting from the reactant in excess instead of the limiting one.
Choosing the limiting reactant by the smallest mass or the smallest number of moles, without dividing by the coefficients.
Dividing actual by theoretical the wrong way round. A yield is never above 100%.
Rounding the moles too early. Keep three or four significant figures until the end.
Self-check 4.0 g of H₂ (M = 2.0 g/mol) and 16 g of O₂ (M = 32 g/mol) react: 2H₂ + O₂ → 2H₂O. What mass of water (M = 18 g/mol) can form?
18 g
20 g
36 g
9 g
Answer: 18 g. O₂ is limiting: 0.50 mol of O₂ gives 1.0 mol of water, 18 g. 36 g starts from the hydrogen, which is in excess. 20 g adds the masses, but part of the hydrogen does not react.
7How exam questions look
Easy: mass to mass. One reactant is given, the others are in excess. Find the mass of a product.
Medium: limiting reactant. Two reactant masses are given. Find the limiting reactant, then the mass of product.
Hard: percent yield. Calculate the theoretical yield, then compare it with the mass that was collected.
The molar masses are given in the question. Write the road map m → n → n → m next to every answer.
For the reaction 2Mg + O2 → 2MgO, what mass m of MgO forms from 12 g of Mg? Assume the other reactants are in excess.
M(Mg) = 24.31 g/molM(MgO) = 40.3 g/mol
Find m(g)
Show hints
Convert the given mass to moles: n = m/M.
The coefficients give the mole ratio: 2 mol of product for every 2 mol of reactant.
n(Mg) = 12/24.31 = 0.4936 mol.
Show the answer and the step-by-step solution
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Moles of reactant
n = 12/24.31 = 0.4936 mol
Mole ratio
n(MgO) = 0.4936·2/2 = 0.4936 mol
Mass of product
m = 0.4936·40.3 = 19.89 g
Answer
m = 19.89 g
Example 2Medium
6 g of C3H8 and 20 g of O2 react according to C3H8 + 5O2 → 3CO2 + 4H2O. What mass m of H2O can form?
M(C3H8) = 44.1M(O2) = 32M(H2O) = 18.02 g/mol
Find m(g)
Show hints
Convert both masses to moles.
Divide each by its coefficient: the smaller result belongs to the limiting reactant.
n(C3H8)/1 = 0.1361, n(O2)/5 = 0.125.
Show the answer and the step-by-step solution
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Moles of each reactant
n(C3H8) = 6/44.1 = 0.1361 mol
n(O2) = 20/32 = 0.625 mol
Limiting reactant
0.1361/1 = 0.1361, 0.625/5 = 0.125 → O2
Mass of product
n(H2O) = 0.125·4 = 0.5 mol
m = 0.5·18.02 = 9.01 g
Answer
m = 9.01 g
Example 3Hard
In the reaction CH4 + 2O2 → CO2 + 2H2O, 10 g of CH4 reacts completely (the other reactants are in excess) and 17.01 g of CO2 is collected. Find the percent yield Y.
M(CH4) = 16.04 g/molM(CO2) = 44.01 g/molactual: 17.01 g
Find Y(%)
Show hints
First find the theoretical yield: the mass that would form if everything reacted.
Percent yield = actual/theoretical × 100.
theoretical: 0.6234 mol·1/1·44.01.
Show the answer and the step-by-step solution
Stoichiometry: a balanced equation counts particles, not grams. Convert mass to moles with n = m/M, use the coefficients of the equation as the mole ratio, then convert back with m = n·M. With two reactants, the limiting reactant is the one that runs out first and fixes how much product forms.
Theoretical yield
n(CH4) = 10/16.04 = 0.6234 mol
m = 0.6234·1/1·44.01 = 27.44 g
Percent yield
Y = 17.01/27.44·100 = 62%
Answer
Y = 62%
Questions and answers
What is Stoichiometry?
A chemical equation is a recipe. 2H₂ + O₂ → 2H₂O says that two molecules of hydrogen react with one molecule of oxygen to give two molecules of water.
The numbers in front of the formulas are the coefficients. They count particles, not grams. The equation must be balanced: every atom appears the same number of times on both sides.
How do I solve Stoichiometry problems step by step?
Write and balance the chemical equation.
Convert the given mass to moles: n = m/M.
Use the coefficients as the mole ratio to find the moles of the substance you want.
Convert back to mass with m = n·M. With two reactants, find the limiting reactant first.
What are common mistakes in Stoichiometry?
Using the mass ratio instead of the mole ratio.
Forgetting to balance the equation.
Not checking which reactant is limiting.
What do the Stoichiometry exercises cover?
Easy: mass to mass
Medium: limiting reactant
Hard: percent yield
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.