# Question d9a3e

Mar 18, 2017

$\text{6.25 g}$

#### Explanation:

The balanced chemical equation

${\text{H"_ 2"SO"_ (4(aq)) + "Zn"("OH")_ (2(s)) -> "ZnSO"_ (4(aq)) + 2"H"_ 2"O}}_{\left(l\right)}$

tells you that the two reactants react in a $1 : 1$ mole ratio and that both of them produce zinc sulfate in a $1 : 1$ mole ratio.

This implies that the theoretical yield of zinc sulfate, which represents the maximum amount of zinc sulfate that can be produced given matching numbers of moles of the two reactants, will be determined by the number of moles of sulfuric acid.

This is the case because zinc hydroxide is said to be in excess.

So, use the molar mass of sulfuric acid to convert the sample from grams to moles

3.80 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"SO"_4)/(98.08 color(red)(cancel(color(black)("g")))) = "0.03874 moles H"_2"SO"_4

You can thus say that the reaction will theoretically produce $0.03874$ moles of zinc sulfate. Use the compound's molar mass to convert this to grams

$0.03874 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles ZnSO"_4))) * "161.44 g"/(1color(red)(cancel(color(black)("mole ZnSO"_4)))) = color(darkgreen)(ul(color(black)("6.25 g}}}}$

The answer is rounded to three sig figs.

To find the percent yield of the reaction, simply divide the actual yield, which is what you get when you perform the reaction, by the theoretical yield and multiply the result by 100%

"% yield" = (5.16 color(red)(cancel(color(black)("g"))))/(6.25color(red)(cancel(color(black)("g")))) * 100% = 82.6%#