# 50g of pure CaCO_3 is heated, liberated CO_2 reacted 0.4 mol of moist ammonia to yield only (NH_4_2CO_3. Find the volume of 3 CO_2 left after the reaction at STP?

## A) zero B) $6.72 L$ C) $2.24 L$ D)indeterminate from this data

Dec 4, 2017

6.72L

#### Explanation:

Chemical reaction: 2$N {H}_{3} + + C {O}_{2} \Rightarrow N {H}_{2} C O O N {H}_{4}$

NH3 reacts with CO2 to give Ammonium carbamate.

Moles of $C {O}_{2}$ taken:50/100=0.5 mol

Moles of $N {H}_{3}$ reacted$\rightarrow$0.4 mol{given}

According to the reaction:
2 moles of $N {H}_{3}$ react with$\rightarrow$1 mole of$C {O}_{2}$

For the reaction of 0.4 mole of $N {H}_{3} \rightarrow$1/2$\times$0.4 mol of CO2 reacted, $\rightarrow$0.4$\times$0.5 moles of $C {O}_{2}$

moles of $C {O}_{2}$ left after reaction= moles taken$-$moles reacted
$=$0.5$-$0.4$\times$0.5$=$0.5(1$-$0.4)$\Rightarrow$0.6/2 $\Rightarrow$0.3

moles of $C {O}_{2}$ left after reaction $\rightarrow$0.3 moles

Volume of 0.3 moles of $C {O}_{2} =$22.4L$\times$0.3{1 mole$\Rightarrow$22.4L}
$\Rightarrow$6.72L