A sample of impure magnesium carbonate was heated to complete decomposition according to
the equation:
MgCO3(s) → MgO(s) + CO2(g).
The mass of the sample is 29.1009g. After the reaction, the mass of the solid residue (consisting
of MgO and the original impurities) is 17.4084 g. How much magnesium carbonate was present
in the original sample, assuming that only the magnesium carbonate had decomposed?
O 10.0912g
O 15.2703g
O 11.6925g
O 44.8036g
O 22.4018g
Answers
A sample of impure magnesium carbonate was heated to complete decomposition according to
the equation, MgCO3(s) → MgO(s) + CO2(g).
The mass of sample is 29.1009g. after the reaction, the mass of solid residue is 17.4084g.
To find : the mass of magnesium carbonate was present in the original sample.
solution : let mass of impurities = x
so, the actual mass of magnesium carbonate = 29.1009 - x
the mass of solid residue (only consisting MgO) = 17.4084 - x
so the mass of CO₂ formed = (29.1009 - x) - (17.4084 - x) = 11.6925 g
no of moles of CO₂ = 11.6925/44 = 0.2657 mol
from reaction, it is clear that 1 mol of magnesium carbonate evolved 1 mol of CO₂ gas.
so, 0.2657 mol of CO₂ will need 0.2657 mol.
so the no of moles of magnesium carbonate = 0.2657 mol
molar mass of magnesium carbonate = 84.3139 g/mol
so the mass of magnesium carbonate = 84.3139 × 0.2657 = 22.4022 g [nearest value of option (4) ]
Therefore option (4) is correct choice.
Answer:
22.4018 is the correct Answer