SME
General Chemistry I · Problem type
The volume of gas a reaction makes
Use this for “2KClO₃ → 2KCl + 3O₂. What volume of O₂ at 25 °C and 1.00 atm does 5.00 g of KClO₃ make?”
Diagram
n(the
reactant)
=
5
g
122.55
g/mol
=
0.0408
mol.
Mole ratio: n(O₂) = 0.0408 × 3/2 = 0.0612 mol.
V
=
nRT
P
V
=
0.0612
×
0.08206
×
298.15
1
V = 1.497 L
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
Mass of the reactant
example · g
M
Molar mass of the reactant
example · g/mol
n₁
Moles of the reactant
calculated · mol
r
Mole ratio (gas per reactant)
example
n₂
Moles of the gas
calculated · mol
t
Temperature in °C
example · °C
T
Temperature
calculated · K
P
Pressure
example · atm
V
Volume of the gas
calculated · L
Formulas
n₁
=
m
M
0.0408
=
5
122.55
n₂
=
r
×
n₁
0.0612 = 1.5 × 0.0408
T
=
t
+
273.15
298.15 = 25 + 273.15
P
×
V
=
n₂
×
0.08206
×
T
1 × 1.497 = 0.0612 × 0.08206 × 298.15
Assumptions
•
The reactant you weigh is used up completely, and every mole of it makes product by the balanced equation.
•
The mole ratio r is the gas’s coefficient over the reactant’s: 3 mol of O₂ for each 2 mol of KClO₃ gives 3/2. Read r from your own balanced equation.
•
The gas is ideal. R = 0.08206 L·atm/(mol·K), so the pressure goes in atm, the volume comes out in liters and the temperature is in kelvins.
Step-by-step
WE KNOW
m = 5 g, M = 122.55 g/mol, r = 1.5, t = 25 °C, P = 1 atm
FIND
moles of the reactant (n₁), moles of the gas (n₂), temperature (T), volume of the gas (V)
Step 1 · Find moles of the reactant (n₁)
Use n₁ = m ÷ M
Grams over the molar mass counts the moles of the reactant.
n₁
=
5
122.55
n₁
=
0.0408
mol
Step 2 · Find moles of the gas (n₂)
Use n₂ = r × n₁
The mole ratio from the balanced equation carries the moles from the reactant to the gas.
n₂
=
1.5
×
0.0408
n₂
=
0.0612
mol
Step 3 · Find temperature (T)
Use T = t + 273.15
The gas laws use kelvins, which start 273.15 degrees below 0 °C.
T
=
25
+
273.15
T
=
298.15
K
Step 4 · Find volume of the gas (V)
Use P × V = n₂ × 0.08206 × T
The
ideal
gas
law
for
the
gas
alone:
V
=
nRT
P
,
in
liters.
V
=
n₂
×
0.08206
×
T
P
V
=
0.0612
×
0.08206
×
298.15
1
V
=
0.0050221
×
298.15
V
=
1.497
L
Check
Put the numbers back in. Both sides should match.
0.0408
=
5
122.55
✓
0.0612 = 1.5 × 0.0408 ✓
298.15 = 25 + 273.15 ✓
1 × 1.497 = 0.0612 × 0.08206 × 298.15 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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