Using the molar volume in calculations with chemical equations
5 marks · HardWhen sodium nitrate (ce{NaNO3}) decomposes on heating, it produces sodium nitrite (ce{NaNO2}) and oxygen gas (ce{O2}). The equation is:
Calculate the number of moles in 0.34 g of sodium nitrate. (Ar: Na = 23, N = 14, O = 16)
Use your answer to (a) to calculate the volume, in cm³, of oxygen gas produced at rtp.
(Molar volume = 24,000 cm³ at rtp.)
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) 0.004 mol
(b) 48 cm³
Common mistakes: Incorrect relative formula mass; Wrong use of mole ratio; Using molar volume in dm³ instead of cm³
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Correctly calculates Mr of NaNO₃: 23 + 14 + (3×16) = 85
A1 (a) · 1 mark · Moles = 0.34 / 85 = 0.004 (allow 0.0040 to 0.0041)
M2 (b) · 1 mark · Uses correct mole ratio O₂:NaNO₃ (2:2 or 1:2); moles O₂ = 0.002 from answer to (a) / 2. (follow-through)
M3 (b) · 1 mark · Uses volume = moles × molar volume in cm³: 0.002 × 24,000. Allow ecf from (a). (follow-through)
A2 (b) · 1 mark · Final answer: 48 cm³ (accept 48.0, or answer from ECF). (follow-through)
Worked solution
(a) Calculate Mr of NaNO₃:
Number of moles = mol
(b) Use the equation: 2 NaNO₃ → 2 NaNO₂ + O₂
Mole ratio: 2 NaNO₃ produces 1 O₂, so moles O₂ = 0.004 / 2 = 0.002
Volume of O₂ (in cm³) = 0.002 × 24,000 = 48 cm³
Final answer: 48 cm³