SME
Grade 10 Science · Problem type
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?”
Diagram
28.01
g
28.01
g/mol
=
1
mol
N₂
1 mol × 2/1 = 2 mol NH₃
5.05
g
2.02
g/mol
=
2.5
mol
H₂
2.5 mol × 2/3 = 1.667 mol NH₃
H₂ makes less NH₃, so H₂ runs out first: it is the limiting reactant.
NH₃ made: 1.667 mol × 17.03 g/mol = 28.38 g.
This is an example. Type your own values to start a new problem.
m₁
Mass of N₂
example · g
m₂
Mass of H₂
example · g
n₁
Moles of N₂
worked out (not typed) · mol
n₂
Moles of H₂
worked out (not typed) · mol
y₁
NH₃ the N₂ could make
worked out (not typed) · mol
y₂
NH₃ the H₂ could make
worked out (not typed) · mol
n
Moles of NH₃ made
worked out (not typed) · mol
m
Mass of NH₃ made
worked out (not typed) · g
x
Excess reactant left over
worked out (not typed) · g
Formulas
n₁
=
m₁
28.01
1
=
28.01
28.01
n₂
=
m₂
2.02
2.5
=
5.05
2.02
y₁
=
n₁
×
2
2 = 1 × 2
y₂
=
n₂
×
2
3
1.67
=
2.5
×
2
3
n
=
the
smaller
of
y₁
and
y₂
1.67 = the smaller of 2 and 1.67
m
=
n
×
17.03
28.4 = 1.67 × 17.03
x
=
m₁
−
n
2
×
28.01
4.67 = 28.01 − 1.67/2 × 28.01
Assumptions
•
The equation is balanced: N₂ + 3 H₂ → 2 NH₃.
•
Grams can’t be compared directly: change each to moles, then to the product’s moles.
•
The reactant that makes less product runs out first: it is the limiting reactant.
Step-by-step
WE KNOW
m₁ = 28.01 g, m₂ = 5.05 g
FIND
moles of N₂ (n₁), moles of H₂ (n₂), NH₃ the N₂ could make (y₁), NH₃ the H₂ could make (y₂), moles of NH₃ made (n), mass of NH₃ made (m), excess reactant left over (x)
Step 1 · Find moles of N₂ (n₁)
Use n₁ = m₁ ÷ 28.01
Divide the grams of N₂ by its molar mass, 28.01 g/mol.
n₁
=
28.01
28.01
n₁
=
1
mol
Step 2 · Find moles of H₂ (n₂)
Use n₂ = m₂ ÷ 2.02
Divide the grams of H₂ by its molar mass, 2.02 g/mol.
n₂
=
5.05
2.02
n₂
=
2.5
mol
Step 3 · Find NH₃ the N₂ could make (y₁)
Use y₁ = n₁ × 2
Mole ratio: 1 N₂ makes 2 NH₃.
y₁
=
1
×
2
y₁
=
2
mol
Step 4 · Find NH₃ the H₂ could make (y₂)
Use y₂ = n₂ × 2/3
Mole ratio: 3 H₂ make 2 NH₃.
y₂
=
2.5
×
2
3
y₂
=
1.6667
mol
Step 5 · Find moles of NH₃ made (n)
Use n = the smaller of y₁ and y₂
The reaction stops when the limiting reactant runs out, so only the smaller amount forms.
n
=
the
smaller
of
2
and
1.6667
1.67 mol < 2 mol: H₂ runs out first, so H₂ is the limiting reactant
n
=
1.6667
mol
Step 6 · Find mass of NH₃ made (m)
Use m = n × 17.03
Multiply the moles of NH₃ by its molar mass, 17.03 g/mol.
m
=
1.6667
×
17.03
m
=
28.4
g
Step 7 · Find excess reactant left over (x)
Use x = m₁ − n/2 × 28.01
N₂ could make more NH₃, so it is in excess: half the moles of NH₃ made is the N₂ used, at 28.01 g/mol.
2 > 1.6667: N₂ in excess
x
=
28.01
−
1.6667/2
×
28.01
x
=
28.01
−
0.83335
×
28.01
x
=
28.01
−
23.342
x
=
4.67
g
Check
Put the numbers back in. Both sides should match.
1
=
28.01
28.01
✓
2.5
=
5.05
2.02
✓
2 = 1 × 2 ✓
1.6667
≈
2.5
×
2
3
✓
1.6667 = the smaller of 2 and 1.6667 ✓
28.4 ≈ 1.6667 × 17.03 ✓
4.67 ≈ 28.01 − 1.6667/2 × 28.01 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.