8.96 L (at STP) of a CO and $CO X_{2}$ mixture has an average molar mass of 36 g/mol. It is bubbled through excess lime water. Taking molar volume 22.4 L/mol and C = 12, O = 16, Ca = 40, the mass of $CaCO X_{3}$ precipitated is

Answer: B

Total moles = 8.96/22.4 = 0.4. Let the mole fraction of CO be x: 28x + 44(1 - x) = 36, so x = 0.5.

CO X_{2}

= 0.2 mol. CO does not react with lime water, and each

CO X_{2}

gives one

CaCO X_{3}

(100 g/mol), so mass = 0.2 x 100 = 20.0 g.