Question

The mass of AgCl precipitated when a solution containing 11.70 g of NaCl is added to a solution containing 3.4g of $AgN{O}_3$ is:
[Atomic mass of Ag = 108, Atomic mass of Na = 23]

• A. 5.74 g
• B. 1.17 g
• C. 2.87 g
• D. 6.8 g

Answer 1

The correct option is C 2.87 g

$NaCl+AgN{O}_3\to AgCl+NaN{O}_3$
$M_{AgCl}=143.5u$, $M_{NaCl}=58.5u$, $M_{AgN{O}_3}=170u$
Moles of $NaCl=\frac{11.70}{58.5}=0.2mol$
Moles of $AgN{O}_3\frac{3.4}{170}=0.02mol$
$AgN{O}_3$ is the limiting reagent here, all of it gets used up.
Mass of AgCl = $0.02\times 143.5=2.87g$