Question

according to the following reaction, how many grams of zinc hydroxide are required for the complete reaction of 32.0 grams of sulfuric acid? sulfuric acid (aq) + zinc hydroxide (s) → zinc sulfate (aq) + water (l) mass = grams

Explanation:

Step1: Write the balanced chemical equation

\[H_2SO_4(aq)+Zn(OH)_2(s)
ightarrow ZnSO_4(aq) + 2H_2O(l)\]

Step2: Calculate the molar mass of \(H_2SO_4\) and \(Zn(OH)_2\)

The molar mass of \(H_2SO_4\): \(M_{H_2SO_4}=2\times1 + 32+4\times16=98\space g/mol\). The molar mass of \(Zn(OH)_2\): \(M_{Zn(OH)_2}=65.38+(16 + 1)\times2=99.38\space g/mol\)

Step3: Calculate the moles of \(H_2SO_4\)

The mass of \(H_2SO_4\) is \(m_{H_2SO_4} = 32.0\space g\). The number of moles of \(H_2SO_4\), \(n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{32.0\space g}{98\space g/mol}\approx0.3265\space mol\)

Step4: Determine the mole - ratio from the balanced equation

From the balanced equation, the mole - ratio of \(H_2SO_4\) to \(Zn(OH)_2\) is \(1:1\). So the moles of \(Zn(OH)_2\) required, \(n_{Zn(OH)_2}=n_{H_2SO_4}= 0.3265\space mol\)

Step5: Calculate the mass of \(Zn(OH)_2\)

The mass of \(Zn(OH)_2\), \(m_{Zn(OH)_2}=n_{Zn(OH)_2}\times M_{Zn(OH)_2}=0.3265\space mol\times99.38\space g/mol\approx32.4\space g\)

Answer:

32.4