Question bank/JEE Main/Chemistry

81 gm mixture of MgCO_3(s) and NH_2COONH_4(s) is heated to
constant mass.
MgCO_3(s) [IMAGE] MgO(s) + CO_2(g)
NH_2COONH_4(s) [IMAGE] 2NH_3(g) + CO_2(g)
If vapour density of gaseous mixture evolved was found to
be$\frac{61}{4}$then find mole of MgCO_3 in theoriginal sample.

A50B60C75DNone of these

Show answer and solution▾

Let no. of moles of MgCO_3 = a
let no. of moles of NH_2COONH_4 = b
MgCO_3(s) [IMAGE] MgO(s) + CO_2(g) a - -
- - a
NH_2COONH_4(s) [IMAGE] 2NH_3(g) + CO_2(g)
b - -
- 2b b
a × 84 + b × 75 = 81.......(1)
$\frac{(a + b) + 4 + (2b)17}{a + b + 2b} = \frac{61}{4} \times 2$........(2)
$a\ = b = \frac{1}{2}$ MgCO_3 = 50

Source: Grass-G / CH-05 Chemistry Paper 10 October.docx