Question

What amount of silver chloride will be obtained when $20mLN/20HCl$ is made to react with excess of $AgN{O}_3$?

Answer 1

we have,

$AgN{O}_3+HCl\to AgCl+HN{O}_3$

so, 1 mole of HCl produce 1 mole of AgCl in presence of excess $AgN{O}_3$

given: 20 ml of $N/20$ HCl

that means,

1000ml of HCl have $1/20$ moles

so in 20 ml we have no. of moles = $\frac{20}{1000\ast 20}$ = 0.001 moles

no. of moles of silver chloride produce = 0.001

amount = no. of moles $\ast$ molecular weight

amount of silver chloride produced = 0.001$\ast$ 143.5 = 0.1435 g