Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

The volume of $CO_2$ obtained by the complete decomposition of $9.85\text{ g}$ of $BaCO_3$ is:

2.24 L

1.12 L

0.84 L

0.56 L

Write the balanced chemical equation: $BaCO_3(s) \xrightarrow{\Delta} BaO(s) + CO_2(g)$
Calculate the Molar Mass of $BaCO_3$ : Atomic masses: $Ba = 137$ , $C = 12$ , $O = 16$ . $M = 137 + 12 + (3 \times 16) = 137 + 12 + 48 = 197\text{ g/mol}$
Calculate moles of $BaCO_3$ : $n = \frac{\text{Given Mass}}{\text{Molar Mass}} = \frac{9.85\text{ g}}{197\text{ g/mol}} = 0.05\text{ mol}$
Apply Stoichiometry: From the equation, $1\text{ mol}$ of $BaCO_3$ produces $1\text{ mol}$ of $CO_2$ . Therefore, $0.05\text{ mol}$ of $BaCO_3$ will produce $0.05\text{ mol}$ of $CO_2$ .
Calculate Volume at STP: At STP, $1\text{ mol}$ of any ideal gas occupies $22.4\text{ L}$ . $V = n \times 22.4\text{ L} = 0.05 \times 22.4 = 1.12\text{ L}$

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