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Chemistry Question on Mole concept and Molar Masses

A sample of CaCO3\text{CaCO}_3 and MgCO3\text{MgCO}_3 weighed 2.21 g is ignited to constant weight of 1.152 g. The composition of the mixture is:
(Given molar mass in g mol−1^{-1})

A

1.187 g CaCO3+1.023 g MgCO31.187 \, \text{g} \, \text{CaCO}_3 + 1.023 \, \text{g} \, \text{MgCO}_3

B

1.023 g CaCO3+1.023 g MgCO31.023 \, \text{g} \, \text{CaCO}_3 + 1.023 \, \text{g} \, \text{MgCO}_3

C

1.187 g CaCO3+1.187 g MgCO31.187 \, \text{g} \, \text{CaCO}_3 + 1.187 \, \text{g} \, \text{MgCO}_3

D

1.023 g CaCO3+1.187 g MgCO31.023 \, \text{g} \, \text{CaCO}_3 + 1.187 \, \text{g} \, \text{MgCO}_3

Answer

1.187 g CaCO3+1.023 g MgCO31.187 \, \text{g} \, \text{CaCO}_3 + 1.023 \, \text{g} \, \text{MgCO}_3

Explanation

Solution

Reactions:

CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g) MgCO3(s)→MgO(s)+CO2(g)\text{MgCO}_3(s) \rightarrow \text{MgO}(s) + \text{CO}_2(g)

Let the weight of CaCO3\text{CaCO}_3 be xx g. Then, the weight of MgCO3\text{MgCO}_3 is (2.21−x)(2.21 - x) g.

Moles of CaCO3\text{CaCO}_3 decomposed = moles of CaO\text{CaO} formed

x100=moles of CaO formed.\frac{x}{100} = \text{moles of $\text{CaO}$ formed.}

Weight of CaO\text{CaO} formed:

weight of CaO formed=x100×56.\text{weight of $\text{CaO}$ formed} = \frac{x}{100} \times 56.

Moles of MgCO3\text{MgCO}_3 decomposed = moles of MgO\text{MgO} formed

2.21−x84=moles of MgO formed.\frac{2.21 - x}{84} = \text{moles of $\text{MgO}$ formed.}

Weight of MgO\text{MgO} formed:

weight of MgO formed=2.21−x84×40.\text{weight of $\text{MgO}$ formed} = \frac{2.21 - x}{84} \times 40.

According to the problem:

2.21−x84×40+x100×56=1.152.\frac{2.21 - x}{84} \times 40 + \frac{x}{100} \times 56 = 1.152.

Solving, we find:

x=1.1886 g (weight of CaCO3).x = 1.1886 \, \text{g (weight of $\text{CaCO}_3$)}.

And the weight of MgCO3\text{MgCO}_3 is:

2.21−x=1.0214 g.2.21 - x = 1.0214 \, \text{g}.