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JEE Advanced levelSingle correct optionDifficulty 3 of 3pb039

$3.7 g$ of $Ca (OH) X_{2}$ is present in water as lime water. $CO X_{2}$ is bubbled in until the initially formed milkiness just disappears. Taking $Ca = 40$ , $C = 12$ , $O = 16$ , $H = 1$ , the maximum mass of $CaCO X_{3}$ formed and the total mass of $CO X_{2}$ used are respectively

A $5.0 g$ and $2.2 g$
B $10.0 g$ and $4.4 g$
C $5.0 g$ and $4.4 g$
D $2.5 g$ and $4.4 g$

Answer: C

n (Ca (OH) X_{2}) = 3.7/74 = 0.05 mol

. First

Ca (OH) X_{2} + CO X_{2} CaCO X_{3} + H X_{2} O

gives

0.05 mol

CaCO X_{3} = 5.0 g

(maximum) using

0.05 mol

CO X_{2}

. Then

CaCO X_{3} + CO X_{2} + H X_{2} O Ca (HCO X_{3}) X_{2}

needs another

0.05 mol

. Total

CO X_{2} = 0.10 \times 44 = 4.4 g

.

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