# How many grams of silver chloride can be produced if you start with "4.62 g" of barium chloride?

## 2"AgNO"_3(aq) + "BaCl"_2(aq) -> 2"AgCl"(s) + "Ba"("NO"_3)_2(aq)

Mar 29, 2018

$\text{6.36 g AgCl}$

#### Explanation:

The balanced chemical equation that describes this double replacement reaction

2"AgNO"_ (3(aq)) + "BaCl"_ (2(aq)) -> 2"AgCl"_ ((s)) darr + "Ba"("NO"_ 3)_ (2(aq))

tells you that when $1$ mole of barium chloride is consumed by the reaction, $2$ moles of silver chloride are produced. Since barium chloride and silver chloride have a $1 : 2$ mole ratio, you can expect the number of moles of barium chloride consumed by the reaction and the number of moles of silver chloride produced by the reaction to be in a $1 : 2$ ratio.

In other words, you have

$\left(\text{moles of BaCl"_2 quad "consumed")/("moles of AgCl produced}\right) = \frac{1}{2}$

Now, in order to be able to use this $1 : 2$ mole ratio, you need to convert the mass of barium chloride to moles. To do that, use the molar mass of barium chloride.

4.62 color(red)(cancel(color(black)("g"))) * "1 mole BaCl"_2/(208.23 color(red)(cancel(color(black)("g")))) = "0.02219 moles BaCl"_2

You can now say that the reaction will produce

0.02219 color(red)(cancel(color(black)("moles BaCl"_2))) * "2 moles AgCl"/(1color(red)(cancel(color(black)("mole BaCl"_2)))) = "0.04438 moles AgCl"

Finally, to convert the number of moles of silver chloride to grams, use the molar mass of the compound.

$0.04438 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles AgCl"))) * "143.32 g"/(1color(red)(cancel(color(black)("mole AgCl")))) = color(darkgreen)(ul(color(black)("6.36 g}}}}$

The answer is rounded to three sig figs.