Question

Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?

Answer 1

Silver nitrate and aluminum chloride react with each other by exchanging anions: 3agno3 (aq) + alcl3 (aq) → al(no3)3 (aq) + 3agcl (s) what mass in grams of agcl is produced when 4.22 g of agno3 react with 7.73 g of alcl3?

Answer 2

Answer: 3.6 grams

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of silver nitrate}=\frac{4.22g}{170g/mol}=0.025moles$

$\text{Number of moles of aluminium chloride}=\frac{7.73g}{133g/mol}=0.06moles$

$3AgNO_3(aq)+AlCl_3(aq)\rightarrow Al(NO)_3)_3(aq)+3AgCl(s)$

According to stoichiometry:

3 mole of $AgNO_3$ react with 1 mole of $AlCl_3$

0.025 moles of $AgNO_3$ react with =$\frac{1}{3}\times 0.025=8.3\times 10^{-3}moles$ of $AlCl_3$

Thus $AgNO_3$ is the limiting reagent as it limits the formation of product and $AlCl_3$ is the excess reagent.

3 moles of $AgNO_3$ produce = 3 moles of $AgCl$

0.025 moles of $AgNO_3$ produce =$\frac{3}{3}\times 0.025=0.025moles$ of $AgCl$

Mass of $AgCl$ produced=$moles\times {\text {Molar mass}}=0.025\times 143=3.6g$

Thus 3.6 g of $AgCl$ is produced.