# Does anyone know how to solve this problem?

## According to the following reaction, how many grams of silver chloride will be formed upon the complete reaction of 24.6 grams of silver nitrate with excess sodium chloride? sodium chloride (aq) + silver nitrate (aq) silver chloride (s) + sodium nitrate (aq) grams silver chloride

May 2, 2018

$20.8$ $g$ ($A g C l$)

#### Explanation:

You'll want to start with a balanced equation.

$N a C l + A g N {O}_{3} \rightarrow A g C l + N a N {O}_{3}$

The molar mass of $A g N {O}_{3}$ is $169.78$ $g m o {l}^{-} 1$

The molar mass of $A g C l$ is $143.32$ $g m o {l}^{-} 1$

Using stoichiometry, we want to find how many moles of silver nitrate is 24.6 grams, then convert moles of silver nitrate to moles of silver chloride, and moles of silver chloride to grams.

Note that, in the balanced equation, there is a 1:1 relationship: one mole of $A g N {O}_{3}$ yields one mole of $A g C l$.

"24.6g" (AgNO_3) * ("1 mol" (AgNO_3))/("169.87g" (AgNO_3)) =0.145 mol

$0.145$ mol$\cdot \left(\text{143.32g" (AgCl))/("1 mol}\right) = 20.8$ $g$ $\left(A g C l\right)$