# How many grams of silver chloride are produced when "1.92 L" of a "0.287-M" silver nitrate solution reacts with enough magnesium chloride?

Sep 27, 2015

$\text{79.0 g}$

#### Explanation:

So, you know that you're dealing with a double replacement reaction between a solution of silver nitrate, ${\text{AgNO}}_{3}$, and solution of magnesium chloride, ${\text{MgCl}}_{2}$

2"AgNO"_text(3(aq]) + "MgCl"_text(2(aq]) -> 2"AgCl"_text((s]) darr + "Mg"("NO"_3)_text(2(aq])

Notice that you have a $1 : 1$ ($2 : 2$) mole ratio between silver nitrate and silver chloride. This tells you that the reaction will rpoduce the number of moles of silver chloride as the number of moles of silver nitrate that reacted.

In your case, you know the molarity and the volume of the silver nitrate solution, which means that you an find the number of moles that react

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A g N {O}_{3}} = \text{0.287 M" * "1.92 L" = "0.55104 moles}$

The aforementioned mole ratio tells you that you get

0.55104color(red)(cancel(color(black)("moles AgNO"""_3))) * "1 mole AgCl"/(1color(red)(cancel(color(black)("mole AgNO"""_3)))) = "0.55104 moles AgCl"

Use silver chloride's molar mass to get the number of grams that would contain this many moles

0.55104color(red)(cancel(color(black)("moles AgCl"))) * "143.32 g"/(color(red)(cancel(color(black)("moleAgCl")))) = "78.975 g"

Rounded to three sig figs, the answer will be

${m}_{A g C l} = \textcolor{g r e e n}{\text{79.0 g}}$