Question

for the following reaction, 4.11 grams of zinc oxide are mixed with excess water. the reaction yields 4.73 grams of zinc hydroxide
zinc oxide (s) + water (l) → zinc hydroxide (aq)
what is the theoretical yield of zinc hydroxide? grams
what is the percent yield for this reaction? %

Explanation:

Step1: Balance the chemical equation

$$ZnO(s)+H_2O(l)\to Zn(OH)_2(aq)$$
The equation is already balanced.

Step2: Calculate the molar mass of \(ZnO\) and \(Zn(OH)_2\)

The molar mass of \(ZnO\): \(M_{ZnO}=65.38 + 16.00=81.38\space g/mol\)
The molar mass of \(Zn(OH)_2\): \(M_{Zn(OH)_2}=65.38+(16.00 + 1.01)\times2=99.40\space g/mol\)

Step3: Calculate the moles of \(ZnO\)

$$n_{ZnO}=\frac{m_{ZnO}}{M_{ZnO}}=\frac{4.11\space g}{81.38\space g/mol}\approx0.0505\space mol$$

Step4: Calculate the theoretical moles and mass of \(Zn(OH)_2\)

From the balanced equation, \(n_{Zn(OH)_2}=n_{ZnO} = 0.0505\space mol\)
$$m_{Zn(OH)_2}^{theoretical}=n_{Zn(OH)_2}\times M_{Zn(OH)_2}=0.0505\space mol\times99.40\space g/mol\approx5.02\space g$$

Step5: Calculate the percent yield

$$Percent\space yield=\frac{m_{Zn(OH)_2}^{actual}}{m_{Zn(OH)_2}^{theoretical}}\times100\%=\frac{4.73\space g}{5.02\space g}\times100\%\approx94.2\%$$

Answer:

The theoretical yield of zinc hydroxide is \(5.02\) grams.
The percent yield for this reaction is \(94.2\%\)