Question

Raghav was performing an experiment in which he combined 0.5 g of ammonia with 1.1 g of HCl to produce ammonium chloride. Again he reacted the ammonium chloride produced with 5.2 g of silver nitrate and formed X g of ammonium nitrate and 4.4 g of silver chloride precipitates out. Find the value of X.

• A. 9.8 gm
• B. 1.2 gm
• C. 10.8 gm
• D. 11.2 gm

Answer 1

The correct option is C

10.8 gm

According to law of conservation of mass:
The reactions are as follows:
$N{H}_3\left(aq\right)+HCl\left(aq\right)\to N{H}_{4}Cl\left(aq\right)$
0.5g 1.1g 1.6g

$N{H}_{4}Cl\left(aq\right)+AgN{O}_3\left(aq\right)\to N{H}_{4}N{O}_3\left(aq\right)+AgCl\downarrow$
1.6g 5.2g X g 4.4g

$1.6g+5.2g=Xg+4.4g$
$X=(1.6+5.2)-4.4=2.4g$