Question

`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) `6.72L`
C) `2.24L`
D)indeterminate from this data

Answer 1

6.72L

Explanation:

Chemical reaction: 2`NH_3 + + CO_2 rArr NH_2COONH_4`

NH3 reacts with CO2 to give Ammonium carbamate.

Moles of `CO_2` taken:50/100=0.5 mol

Moles of `NH_3` reacted`rarr`0.4 mol{given}

According to the reaction:
2 moles of `NH_3` react with`rarr`1 mole of`CO_2`

For the reaction of 0.4 mole of `NH_3 rarr`1/2`xx`0.4 mol of CO2 reacted, `rarr`0.4`xx`0.5 moles of `CO_2`

moles of `CO_2` left after reaction= moles taken`-`moles reacted
`=`0.5`-`0.4`xx`0.5`=`0.5(1`-`0.4)`rArr`0.6/2 `rArr`0.3

moles of `CO_2` left after reaction `rarr`0.3 moles

Volume of 0.3 moles of `CO_2=`22.4L`xx`0.3{1 mole`rArr`22.4L}
`rArr`6.72L