# Question ac8d7

Sep 26, 2017

74.1%

#### Explanation:

Start by writing the balanced chemical equation that describes this double-replacement reaction

${\text{Na"_ 3"PO"_ (4(aq)) + "AlCl"_ (3(aq)) -> "AlPO"_ (4(s)) darr + 3"NaCl}}_{\left(a q\right)}$

Notice that in order to produce $1$ mole of aluminium phosphate, the reaction must consume $1$ mole of sodium phosphate and $1$ mole of aluminium chloride.

You can use the molar masses of the three chemical species

• M_ ("M Na" _ 3"PO"_ 4) = "163.94 g mol"^(-1)
• M_ ("M AlCl"_ 3) = "133.34 g mol"^(-1)
• M_ ("M AlPO" _4) = "121.953 g mol"^(-1)

to say that in order for the reaction to produce $\text{121.953 g}$ of aluminium phosphate, it must consume $\text{163.94 g}$ of sodium phosphate and $\text{133.34 g}$ of aluminium chloride.

Now, you know that aluminium chloride is in excess, which implies that all the mass of sodium phosphate present in your sample will actually react.

This means that at 100% yield, the reaction should produce

33.4 color(red)(cancel(color(black)("g Na"_3"PO"_4))) * "121.953 g AlPO"_4/(163.94color(red)(cancel(color(black)("g Na"_3"PO"_4)))) = "24.846 g AlPO"_4

This represents the reaction's theoretical yield.

However, you know that your reaction produced $\text{18.4 g}$ of aluminium phosphate. This value represents the reaction's actual yield.

To find the reaction's percent yield, simply divide the actual yield by the theoretical yield and multiply the result by 100%.

"% yield" = (18.4 color(red)(cancel(color(black)("g"))))/(24.846color(red)(cancel(color(black)("g")))) * 100% = color(darkgreen)(ul(color(black)(74.1%)))#