# Question 58f81

Nov 27, 2016

$38.418$ $\text{g}$

#### Explanation:

"MgCl"_2+2"AgNO"_3rarr2"AgCl"+"Mg""(NO"_3)_2#

In order to work out the mass of silver chloride, we must first calculate the number of moles we have reacting. This can be done using the following formula:

$\text{moles (n)"=("mass (m)")/"Atomic mass (M)}$

We don't have the mass of silver chloride, but we do the mass of magnesium chloride, so we have to work out the moles of magnesium chloride instead.

moles of $\text{MgCl"_2="m"/"M} = \frac{12.7}{95} = 0.134$

Now that we have the moles of magnesium chloride, we have to use stoichiometry to work out the moles of silver chloride. To do this, we use the molar ratio.

The molar ratio is literally just the ratio of the numbers that come before the reactants and products. So the reacting ratio of magnesium chloride : silver chloride is 1 : 2.

Molar ratio of $\text{MgCl"_2 : "AgCl} = 1 : 2$ so moles of $\text{AgCl} = 0.134 \times 2 = 0.268$

Now that we have the moles of silver chloride, we can work out the expected mass by rearranging the mole equation.

$\text{n"="m"/"M}$

$\text{m"="n"xx"M}$

mass of $\text{AgCl"="n"xx"M} = 0.268 \times 143.5 = 38.418$ $\text{g}$