Question

Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be produced from 5.0 kg of limestone.

Answer 1

Answer: 2800 g

Explanation:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5 kg = 5000 g

$\text{Number of moles}=\frac{5000g}{100g/mol}=50moles$

1 mole of $CaCO_3$ produces = 1 mole of $CaO$

50 moles of $CaCO_3$ produces =$\frac{1}{1}\times 50=50moles$ of $CaO$

Mass of $CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g$

2800 g of $CaO$ is produced from 5.0 kg of limestone.