Question

What mass of silver chloride will be obtained by adding an excess of hydrochloric acid to a solution of 0.34 g of silver nitrate?
[Cl=35.5, Ag=108, N=14, O=16, H=1]

• A. 0.287 g
• B. 0.34 g
• C. 2.87 g
• D. 0.07 g

Answer 1

Answer: (A)

0.287 g

$AgN{O}_3+HCl\to AgCl+HN{O}_3$
1 1 1
170 143.5
Mass of $AgCl$ formed $=143.5\times 0.34/170=0.287gm$