Question

What mass of silver chloride can be produced from 1.32 L of a 0.273 M solution of silver nitrate?

Answer 1

Number of moles of silver nitrate$=Vol \times Molarity=1.32 \times 0.273=0.36036$

Reaction to displace nitrate to chloride:

$AgNO_3+Cl^-\to AgCl+NO_3^-$

so ,Number of moles of silver chloride formed$=$ Number of moles of silver nitrate used=0.36036

Mass of silver chloride formed$=Molar \ mass\times Number\ of \ moles=0.36036 \times 143.32=51.647 gm$