Question

How many grams of silver chloride are formed when 11.0 g of silver nitrate reacts with 14.0 g of barium chloride

Answer 1

$2AgNO_3(aq)+BaCl_2(aq)=Ba(NO_3)_2(aq)+2AgCl(s)$

moles of $AgNO_3 = 11.0 g \frac{1mole}{169.87g}=0.0648 mol$

Moles of $BaCl_2= 14.0 g \frac{1 mole}{208.23 g}= 0.0672 mol$

Find limiting reactant:

Compare $\frac{moles AgNO_3}{2}$ and $\frac{moles BaCl_2}{1}$

$\frac{0.0648}{2}<\frac{0.0672}{1}$ AgNO3 is limiting reactant.

Find mass of AgCl

$0.0648 mol AgNO_3 \frac{ 2 mol AgCl}{2 mol AgNO_3} \frac{143.32 g AgCl}{1 mol AgCl}= 9.29 g AgCl$