Question

4gm of lead nitrate reacts with 1.17g of sodium chloride and form 3gm of lead chloride how much odium nitrate will form?

Answer 1

2NaCl (aq) + Pb(NO₃)₂ (aq) → PbCl₂ (s) + 2NaNO₃ (aq)

$Moles (NaCl)=\frac{1.17}{58.44}=0.02moles$

Ratio of $NaCl$ and $NaNO_3$ is 1:1

Hence

$Moles (NaNO_3)=0.02moles.$ 4gm