1.84 g of a mixture of CaCO_3 and MgCO_3 is | vedclass

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(C) Let the mass of $CaCO_3$ be $x \ g$ and the mass of $MgCO_3$ be $(1.84 - x) \ g$. Heating the carbonates results in the following reactions: $CaCO

Vedclass · Accepted Answer

$54.34$

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(C) Let the mass of $CaCO_3$ be $x \ g$ and the mass of $MgCO_3$ be $(1.84 - x) \ g$. Heating the carbonates results in the following reactions: $CaCO_3(s) ightarrow CaO(s) + CO_2(g)$ $MgCO_3(s) ightarrow MgO(s) + CO_2(g)$ The molar mass of $CaCO_3 = 100 \ g/mol$,$CaO = 56 \ g/mol$,$MgCO_3 = 84 \ g/mol$,and $MgO = 40 \ g/mol$. The mass of the residue $(CaO + MgO)$ is $0.96 \ g$. Mass of $CaO = (x / 100) 	imes 56 = 0.56x$. Mass of $MgO = ((1.84 - x) / 84) 	imes 40 = 0.476(1.84 - x)$. $0.56x + 0.476(1.84 - x) = 0.96$. $0.56x + 0.876 - 0.476x = 0.96$. $0.084x = 0.084$. $x = 1 \ g$. Percentage of $CaCO_3 = (1 / 1.84) 	imes 100 \approx 54.34\%$.

$1.84 \ g$ of a mixture of $CaCO_3$ and $MgCO_3$ is strongly heated to get a residue of $0.96 \ g$. The percentage of $CaCO_3$ in the mixture is

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