# Question #248d8

Feb 11, 2017

Silver nitrate. 9.11g Silver chloride.

#### Explanation:

First convert the given quantities of compounds to moles. Then apply the balanced equation. 10.8g $A g N {O}_{3}$ is
$\left(10.8 \frac{g}{170 \frac{g}{m o l}}\right)$ = 0.0635 moles $A g N {O}_{3}$ .

0.1276L * 0.500mol/L = 0.0638 moles $C a C {l}_{2}$.

So, we have almost equal molar quantities of each reactant, but the reaction equation shows that TWO moles of $A g N {O}_{3}$ are needed to react with ONE mole of $C a C {l}_{2}$. Therefore, the $A g N {O}_{3}$ is the limiting reagent, leaving half of the chloride in solution.

The maximum products that can be formed are ONE mole of AgCl for every ONE mole of $A g N {O}_{3}$ . Because we already determined that it is the limiting reagent, it will be completely reacted, producing 0.0635 moles of silver chloride.

0.0635mol * 143.5g/mol = 9.11g AgCl.