Question

The right option for the mass of CO, produced by heating 20 g of 20% pure limestone is (Atomic mass of Ca = 40):

\[\ce{CaCO3 (at 1200 K) -> CaO + CO2}\]

Options

• 1.76 g

• 2.64 g

• 1.32 g

• 1.12 g

Answer 1

1.76 g

Explanation:

Given: Weight of impure limestone = 20 g

Weight of pure limestone (CaCO₃) = 20% of 20 g

= `20/100 xx 20`

= 4 g

`"n"_("CaCO"_3) = 4/100` = 0.04

\[\ce{\underset{n = 0.04}{CaCO3} -> CaO + \underset{n = 0.04}{CO2}}\]

`"n"_("CO"_2)` = 0.04

`"W"_("CO"_2)` = 0.04 × 44

= 1.76 g