Question

90 g of butane on combustion with 80 g of oxygen gave carbon dioxide and water. The gas formed in the reaction is reacted with 40 g of calcium carbonate. Find the amount of product $\left[Ca\right(HC{O}_3{)}_2]$ formed.
$C_4{H}_{10}+O_2\to C{O}_2+H_{2}OCaC{O}_3+C{O}_2+H_{2}O\to Ca(HC{O}_3{)}_2$
(Molar mass of Ca = 40 g/mol)

• A. 44.5 g
• B. 162.1 g
• C. 64.8 g
• D. 120.8 g

Answer 1

The correct option is C 64.8 g

$2C_4{H}_{10}+13O_2\to 8C{O}_2+10H_{2}O$
$CaC{O}_3+C{O}_2+H_{2}O\to Ca(HC{O}_3{)}_2$
$\text{Moles of}{C}_4{H}_{10}=\frac{\text{given mass}}{\text{molar mass}}=\frac{90}{58}=1.55mol\text{Moles of}{O}_2=\frac{\text{given mass}}{\text{molar mass}}=\frac{80}{32}=2.5mol$
Finding the limiting reagent:
2 moles of butane react with 13 moles of oxygen.
1.55 moles will react with $\frac{13}{2}\times 1.55=10.075mol$
So, oxygen will be the limiting reagent.
13 moles of oxygen produce 8 moles of $C{O}_2$
2.5 moles will produce = $\frac{8}{13}\times 2.5=1.54$ moles of $C{O}_2$
In reaction(ii),
$MolesofCaC{O}_3=\frac{\text{given mass}}{\text{molar mass}}=\frac{40}{100}=0.4mol$
$CaC{O}_3$ will be the limiting reagent.
1 mole of $CaC{O}_3$ produces 1 mole of $Ca(HC{O}_3{)}_2$
So, the amount of $Ca(HC{O}_3{)}_2$ formed = $0.4mol$
Mass of $Ca(HC{O}_3{)}_2$
= 0.4 mol $\times$ 162
= 64.8 g