Question

aqueous sulfuric acid (h₂so₄) reacts with solid sodium hydroxide (naoh) to produce aqueous sodium sulfate (na₂so₄) and liquid water (h₂o). what is the theoretical yield of water formed from the reaction of 2.9 g of sulfuric acid and 0.91 g of sodium hydroxide?
be sure your answer has the correct number of significant digits in it.

Explanation:

Step1: Write the balanced chemical equation

$$\ce{H_{2}SO_{4}(aq) + 2NaOH(s) ightarrow Na_{2}SO_{4}(aq)+2H_{2}O(l)}$$

Step2: Calculate the molar mass of reactants

• Molar mass of $\ce{H_{2}SO_{4}}$: $M_{H_{2}SO_{4}}=(2\times1)+32+(4\times16)=98\space g/mol$
• Molar mass of $\ce{NaOH}$: $M_{NaOH}=23 + 16+1=40\space g/mol$

Step3: Calculate the number of moles of reactants

• Moles of $\ce{H_{2}SO_{4}}$: $n_{H_{2}SO_{4}}=\frac{m_{H_{2}SO_{4}}}{M_{H_{2}SO_{4}}}=\frac{2.9\space g}{98\space g/mol}\approx0.0296\space mol$
• Moles of $\ce{NaOH}$: $n_{NaOH}=\frac{m_{NaOH}}{M_{NaOH}}=\frac{0.91\space g}{40\space g/mol}=0.02275\space mol$

Step4: Determine the limiting reactant

From the balanced equation, the mole ratio of $\ce{H_{2}SO_{4}}$ to $\ce{NaOH}$ is $1:2$.
For $0.0296\space mol$ of $\ce{H_{2}SO_{4}}$, the moles of $\ce{NaOH}$ required is $2\times0.0296 = 0.0592\space mol$. But we have only $0.02275\space mol$ of $\ce{NaOH}$.
For $0.02275\space mol$ of $\ce{NaOH}$, the moles of $\ce{H_{2}SO_{4}}$ required is $\frac{0.02275}{2}=0.011375\space mol$. Since $0.011375\space mol<0.0296\space mol$, $\ce{NaOH}$ is the limiting reactant.

Step5: Calculate the moles of water produced

From the balanced equation, mole ratio of $\ce{NaOH}$ to $\ce{H_{2}O}$ is $2:2 = 1:1$. So moles of $\ce{H_{2}O}$ produced $n_{H_{2}O}=n_{NaOH}=0.02275\space mol$

Step6: Calculate the mass of water

Molar mass of $\ce{H_{2}O}$: $M_{H_{2}O}=(2\times1)+16 = 18\space g/mol$
Mass of $\ce{H_{2}O}$: $m_{H_{2}O}=n_{H_{2}O}\times M_{H_{2}O}=0.02275\space mol\times18\space g/mol = 0.41\space g$

Answer:

$0.41\space g$