The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg.

(Nearest integer)

Given: Molar mass (in g ${mol}^{- 1}$ ) of Ca–40, O–16, C–12 [2025]

Question

The amount of calcium oxide produced on heating 150 kg limestone (75% pure) is _____ kg.

(Nearest integer)

Given: Molar mass (in g ${mol}^{- 1}$ ) of Ca–40, O–16, C–12 [2025]

Smart Achievers · Accepted Answer

(63)

${Mass of CaCO}_{3} in 150 kg sample of limestone (W_{{CaCO}_{3}}) = 75 % of 150 kg$ $= 0.75 \times 150 kg = 112.5 kg = 112500 g$

${Molar mass of CaCO}_{3} (M_{{CaCO}_{3}}) = 100 g/mol$

${Moles of CaCO}_{3} (n_{{CaCO}_{3}}) = \frac{W_{{CaCO}_{3}}}{M_{{CaCO}_{3}}}$

$= \frac{112500}{100} = 1125 mol$

Decomposition of calcium carbonate is represented as:

${CaCO}_{3} (s) \overset{Δ}{\to} CaO (s) + {CO}_{2} (g)$

$By stoichiometry,$

$n_{CaO} = n_{{CaCO}_{3}} = 1125 mol$

$Mass of CaO = n_{CaO} \times M_{CaO}$ $= 1125 \times 56 g = 63000 g = 63 kg$

Want Us to Email you About Special Offers & Updates?