JEE MAIN - Chemistry (2025 - 4th April Evening Shift - No. 22)
The amount of calcium oxide produced on heating 150 kg limestone ( $75 \%$ pure) is _________ kg. (Nearest integer)
Given: Molar mass (in $\mathrm{g} \mathrm{mol}^{-1}$ ) of Ca-40, O-16, C-12
Answer
63
Explanation
$$\begin{aligned} & \mathrm{CaCO}_3 \rightarrow \mathrm{CaO}+\mathrm{CO}_2 \\ & \text { mass of } \mathrm{CaCO}_3=\frac{150 \times 75}{100}=112.5 \mathrm{~kg} \\ & =112500 \mathrm{~g} \\ & \mathrm{n}_{\mathrm{CaCO}_3}=1125 \end{aligned}$$
$$\begin{aligned} & \text { So moles of } \mathrm{CaO}=1125 \\ & \qquad \text { mass of } \mathrm{CaO}=\frac{1125 \times 56}{1000}=63 \mathrm{~kg} \\ & \text { Correct answer } \Rightarrow 63 \end{aligned}$$
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