Question

for the following reaction, 30.3 grams of zinc oxide are allowed to react with 5.69 grams of water.
zinc oxide $(s)+$ water $(l)
ightarrow$ zinc hydroxide $(a q)$
what is the maximum mass of zinc hydroxide that can be formed?
mass $=square mathrm{g}$
what is the formula for the limiting reactant?

what mass of the excess reagent remains after the reaction is complete?
mass $=square mathrm{g}$
the equation for the reaction is:
$mathrm{zno}(s)+mathrm{h}_{2} mathrm{o}(l)
ightarrow mathrm{zn}(mathrm{oh})_{2}(a q)$

Explanation:

Step1: Calculate moles of reactants

Molar mass of \(ZnO\) is \(81.38\ g/mol\). Moles of \(ZnO=\frac{30.3\ g}{81.38\ g/mol}\approx0.372\ mol\).
Molar mass of \(H_{2}O\) is \(18.02\ g/mol\). Moles of \(H_{2}O=\frac{5.69\ g}{18.02\ g/mol}\approx0.316\ mol\).

Step2: Determine limiting reactant

From the balanced equation \(ZnO + H_{2}O
ightarrow Zn(OH)_{2}\), the mole ratio of \(ZnO\) to \(H_{2}O\) is \(1:1\). Since \(0.316\ mol\) (moles of \(H_{2}O\)) \(< 0.372\ mol\) (moles of \(ZnO\)), \(H_{2}O\) is the limiting reactant.

Step3: Calculate moles of \(Zn(OH)_{2}\)

Using the mole ratio \(1:1\) (from balanced equation), moles of \(Zn(OH)_{2}=0.316\ mol\).
Molar mass of \(Zn(OH)_{2}\) is \(99.42\ g/mol\). Mass of \(Zn(OH)_{2}=0.316\ mol\times99.42\ g/mol\approx31.4\ g\).

Step4: Calculate mass of excess \(ZnO\)

Moles of \(ZnO\) reacted \( = 0.316\ mol\). Moles of \(ZnO\) remaining \(=0.372 - 0.316=0.056\ mol\).
Mass of \(ZnO\) remaining \(=0.056\ mol\times81.38\ g/mol\approx4.56\ g\).

Answer:

Mass of \(Zn(OH)_{2}=31.4\ g\)
Formula for limiting reactant: \(H_{2}O\)
Mass of excess reagent (\(ZnO\)) remaining \( = 4.56\ g\)