Question

Raghav performed 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 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

Follow the law of consevation of mass.

The reactions are as follows:
$N{H}_3+HCl\to N{H}_{4}Cl$
$0.5g1.1g1.6g$

$N{H}_{4}Cl+AgN{O}_3\to N{H}_{4}N{O}_3+AgCl$
$1.6g5.2gXg4.4g$

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