Question

What weight of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of $AgN{O}_3$ ?
(Molar mass of Na = 23 g/mol, Cl = 35.5 g/mol, Ag = 108 g/mol)

• A. 3.23 g
• B. 5.23 g
• C. 3.88 g
• D. 4.88 g

Answer 1

The correct option is D 4.88 g

The given balanced chemical equation is,
$AgN{O}_3+NaCl\to AgCl+NaN{O}_3$

Molar mass of NaCl = (23 + 35.5) g/mol = 58.5 g/mol

Molar mass of $AgN{O}_3=(108+48+14)$ g/mol
= 170 g/mol

$4.77gofNaCl=\frac{4.77}{58.5}$ mol
$NaCl$ = 0.082 mol

$5.77gofAgN{O}_3=\frac{5.77}{170}molofAgN{O}_3=0.034mol$

Thus $AgN{O}_3$ is the limiting reagent here.

So, moles of AgCl formed in the reaction = 0.034

Again. Molar mass of AgCl is,
$=(108+35.5)g/mol=143.5g/mol$

Hence, the mass of AgCl produced $=143.5\times 0.034g=4.879g$