SME
Grade 10 Science · Chemistry
Stoichiometry and limiting reactants
Use this for “How many grams of NH₃ form from 6.06 g of H₂?” with the mole ratio from the balanced equation.
Diagram
Mass = 3 mol × 2.02 g/mol = 6.06 g.
Mole
ratio:
3
×
2
3
=
2
mol
of
NH₃.
Every path goes through moles: the filled box is the value given.
This is an example. Type your own values to start a new problem.
m
g
H₂
×
1
mol
H₂
2.02
g
H₂
×
2
mol
NH₃
3
mol
H₂
×
17.03
g
NH₃
1
mol
NH₃
=
q
g
NH₃
n₁
Moles of H₂
calculated · mol
n₂
Moles of NH₃
calculated · mol
Formulas
m
=
2.02
×
n₁
6.06 = 2.02 × 3
n₂
=
n₁
×
2
3
2
=
3
×
2
3
q
=
17.03
×
n₂
34.1 = 17.03 × 2
Assumptions
•
The balanced equation is N₂ + 3H₂ → 2NH₃: 3 mol of H₂ make 2 mol of NH₃.
•
Coefficients count moles, not grams.
•
Always go through moles: grams → moles → mole ratio → moles → grams.
Step-by-step
WE KNOW
m = 6.06 g
FIND
moles of H₂ (n₁), moles of NH₃ (n₂), mass of NH₃ (q)
Step 1 · Find moles of H₂ (n₁)
Use m = 2.02 × n₁
Divide the grams by the molar mass: the grams cancel, leaving moles.
n₁
=
m
/2.02
n₁
=
6.06/2.02
n₁
=
3
mol
Step 2 · Find moles of NH₃ (n₂)
Use n₂ = n₁ × 2/3
The equation makes 2 mol of NH₃ for every 3 mol of H₂.
n₂
=
3
×
2
3
n₂
=
2
mol
Step 3 · Find mass of NH₃ (q)
Use q = 17.03 × n₂
One mole of NH₃ has a mass of 17.03 g.
q
=
17.03
×
2
q
=
34.1
g
Check
Put the numbers back in. Both sides should match.
6.06 = 2.02 × 3 ✓
2
=
3
×
2
3
✓
34.1 ≈ 17.03 × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
More problem types
The limiting reactant
Use this to find which reactant runs out first, how much product forms and what is left over.
The limiting reactant from grams
Use this for “28.01 g of N₂ and 5.05 g of H₂ react. How many grams of NH₃ form?”
Percent yield
Use this for the percent yield from the actual and theoretical masses of product.