A sample of CaCO_3 and MgCO_3 weighing 2.21 | vedclass

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(A) $CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$$MgCO_{3(s)} \xrightarrow{\Delta} MgO_{(s)} + CO_{2(g)}$Let the mass of $CaCO_3$ be $x \ g

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$1.187 \ g \ CaCO_3 + 1.023 \ g \ MgCO_3$

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(A) $CaCO_{3(s)} \xrightarrow{\Delta} CaO_{(s)} + CO_{2(g)}$$MgCO_{3(s)} \xrightarrow{\Delta} MgO_{(s)} + CO_{2(g)}$Let the mass of $CaCO_3$ be $x \ g$.Then the mass of $MgCO_3 = (2.21 - x) \ g$.Mass of $CaO$ formed $= \frac{x}{100} 	imes 56 = 0.56x \ g$.Mass of $MgO$ formed $= \frac{2.21 - x}{84} 	imes 40 = 0.4762(2.21 - x) \ g$.Total mass of residue $= 0.56x + 0.4762(2.21 - x) = 1.152$.$0.56x + 1.0524 - 0.4762x = 1.152$.$0.0838x = 0.0996$.$x \approx 1.188 \ g$ of $CaCO_3$.Mass of $MgCO_3 = 2.21 - 1.188 = 1.022 \ g$.Rounding to the nearest option,the composition is $1.187 \ g \ CaCO_3$ and $1.023 \ g \ MgCO_3$.

$A$ sample of $CaCO_3$ and $MgCO_3$ weighing $2.21 \ g$ is ignited to a constant weight of $1.152 \ g$. The composition of the mixture is:
(Given molar mass in $g \ mol^{-1}: CaCO_3 = 100, MgCO_3 = 84$)

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$A$ sample of $CaCO_3$ and $MgCO_3$ weighing $2.21 \ g$ is ignited to a constant weight of $1.152 \ g$. The composition of the mixture is:(Given molar mass in $g \ mol^{-1}: CaCO_3 = 100, MgCO_3 = 84$)