Question

Calculate the weight of residue obtained when caco3 is strongly heated and 5.6 litre co2 is produced at n.t.p

Answer 1

Answer: 14 grams of residue.

Explanation:

$CaCO_3\rightarrow CaO+CO_2$

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

$\text{Number of moles of carbon dioxide}=\frac{5.6L}{22.4L}=0.25 moles$

According to stoichiometry:

1 mole of $CO_2$ is produced by 1 mole of $CaCO_3$

Thus 0.25 moles of $CO_2$ is produced by =$\frac{1}{1}\times 0.25=0.25moles$ of $CaCO_3$

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

Thus 0.25 moles of $CaCO_3$  produces =$\frac{1}{1}\times 0.25=0.25moles$ of $CaO$  residue

Mass of residue= ${\text{moles}\times {\text{Molar Mass}=0.25\times 56g=14g$

Therefore, the weight of residue obtained is 14 grams.

Answer 2

Answer:

CaCO3(s)⟶CaO(s)+CO2(g)

Moles of CO2 at STP = 5.622.4 = 0.25 mol

1 mol of CaCO3 gives 1 mol of CaO (residue) and 1 mol of CO2.

So, when 0.25 mol of CO2 is produced, 0.25 mol of residue will be obtained

weight of CaO formed =Moles×Molar mass

Hence, weight of CaO=0.25 mol×56 g/mol=14 g