Question

what mass of AgCl will precipitate when 10.0 g of NaCl is added to an aqueous solution of AgNO3?
NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)

Answer 1

Number of moles of NaCl = 10.0 g / 58.44 g/mol = 0.171 mole

From the balanced equation we can say that

1 mole of NaCl produces 1 mole of AgCl so

0.171 mole of NaCl will produce

= 0.171 mole of NaCl *(1 mole of AgCl / 1 mole of NaCl)

= 0.171 mole of AgCl

mass of 1 mole of AgCl = 143.32 g

so the mass of 0.171 mole of AgCl = 24.5 g

Therefore, the mass of AgCl produced would be 24.5 g