Question

How many grams of silver cholride will be precipitated by adding sufficient silver nitrate to react with 1500.0mL of.400M barium chloride solution?

Answer 1

Answer: 172.2 g of $AgCl$ is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}={\text {Molarty}}\times {\text{Volume in L}}=0.400\times 1.5L=0.6moles$

$BaCl_2+2AgNO_3\rightarrow 2AgCl+Ba(NO_3)_2$

According to stoichiometry:

1 mole of $BaCl_2$ produce = 2 moles of $AgCl$

Thus 0.6 moles $BaCl_2$ will produce =$\frac{2}{1}\times 0.6=1.2$ moles of $AgCl$

Mass of $AgCl=moles\times {\text {molar mass}}=1.2mol\times 143.5g/mol=172.2g$

Thus 172.2 g of $AgCl$ is formed by adding sufficient silver nitrate to react with 1500.0mL of 0.400M barium chloride solution.