Carbon dioxide and global warming: the greenhouse effect
2 marks · EasyA student heats 1.2 g of magnesium carbonate (ce{MgCO3}) and all of it decomposes. Calculate the maximum mass of carbon dioxide gas that can be produced. (Relative atomic masses: Mg = 24, C = 12, O = 16)
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0.528 g (±0.01)
Common mistakes: Incorrect use of molar masses; Forgets to use provided relative atomic masses
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Marking points
M1 · 1 mark · Calculates moles of MgCO3: (1.2 / 84) = 0.0143 mol (allows working to use correct molar masses)
A1 · 1 mark · Mass of CO2 = moles x Mr = 0.0143 × 44 = 0.629, answer rounds within 0.01 g to 0.63 g. Accept values due to rounding between 0.62–0.63 g if correct working shown. (follow-through)
Worked solution
Calculate Mr(MgCO3) = 24 + 12 + (16 × 3) = 24 + 12 + 48 = 84
Moles of MgCO3 = 1.2 \div 84 = 0.0143 mol
\ce{MgCO3 -> MgO + CO2}, so moles of CO2 = moles of MgCO3 = 0.0143 mol
Mr(CO2) = 12 + (16 × 2) = 44
Mass of CO2 = 0.0143 × 44 = 0.629 g (rounded, accept ~0.63 g)