Chemical Formulae, Equations and Calculations Part 2
7 marks · HardA student reacts sodium carbonate solution with excess hydrochloric acid at room temperature and pressure (rtp). The equation for the reaction is:
The student wants to produce exactly 2.40 dm³ of carbon dioxide gas measured at rtp.
Calculate the number of moles of carbon dioxide that occupy 2.40 dm³ at rtp.
Using the chemical equation, determine the number of moles of sodium carbonate required to produce this volume of carbon dioxide.
The student has 0.500 mol/dm³ sodium carbonate solution. Calculate the volume of this solution, in cm³, needed to provide the amount of sodium carbonate calculated in part b.
State the conversion factor you would use to convert a volume in dm³ to cm³.
Explain why the molar volume of a gas at rtp is used in this calculation.
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.
Show answer
(a) 0.100 mol
(b) 0.100 mol
(c) 200 cm³ · must show step for volume in dm³, conversion to cm³
(d) Multiply by 1000
(e) Because 1 mole of any gas occupies the same volume at rtp, which allows direct calculation of gas volumes from moles. · must show states molar volume is same for any gas at rtp, allows calculation of volume from moles
Show mark scheme and worked solution
Marking points
A1 (a) · 1 mark · Correct calculation of moles of CO₂ from volume at rtp
A2 (b) · 1 mark · Correct ratio from equation and calculation of sodium carbonate moles (follow-through)
M1 (c) · 1 mark · Rearranges concentration equation or sets up volume calculation (follow-through)
M2 (c) · 1 mark · Converts volume from dm³ to cm³, using correct factor (follow-through)
A3 (c) · 1 mark · Arrives at correct final volume (200 cm³) (follow-through)
B1 (d) · 1 mark · States correct conversion factor (multiply by 1000)
B2 (e) · 1 mark · Explains that 1 mole of any gas occupies same volume at rtp and why this simplifies calculation
Worked solution
a) The molar volume at rtp is 24 dm³ per 1 mol. number of moles of CO₂ = volume / molar volume = 2.40 dm³ ÷ 24 dm³/mol = 0.100 mol
b) From the equation: 1 mol Na₂CO₃ produces 1 mol CO₂. So, 0.100 mol CO₂ needs 0.100 mol Na₂CO₃.
c) Concentration of sodium carbonate = 0.500 mol/dm³. Volume needed:
Convert to cm³: 0.200 dm³ × 1000 = 200 cm³.
d) To convert a volume in dm³ to cm³, multiply by 1000.
e) The molar volume of a gas at rtp is used because, at rtp, 1 mole of any gas occupies the same volume (24 dm³), making it possible to calculate gas volumes directly from the number of moles.