Questions: For the following reaction, 25.2 grams of sulfur dioxide are allowed to react with 5.49 grams of oxygen gas. sulfur dioxide (g)+ oxygen (g) → sulfur trioxide (g) What is the maximum mass of sulfur trioxide that can be formed? Mass = 8 What is the FORMULA for the limiting reactant? What mass of the excess reagent remains after the reaction is complete? Mass = g

Solution Steps

The balanced chemical equation for the reaction is: \[ 2 \text{SO}_2 (g) + \text{O}_2 (g) \rightarrow 2 \text{SO}_3 (g) \]

First, we need to calculate the moles of each reactant.

• Molar mass of \(\text{SO}_2\): \[ \text{Molar mass of SO}_2 = 32.07 + 2 \times 16.00 = 64.07 \, \text{g/mol} \] \[ \text{Moles of SO}_2 = \frac{25.2 \, \text{g}}{64.07 \, \text{g/mol}} = 0.3933 \, \text{mol} \]

• Molar mass of \(\text{O}_2\): \[ \text{Molar mass of O}_2 = 2 \times 16.00 = 32.00 \, \text{g/mol} \] \[ \text{Moles of O}_2 = \frac{5.49 \, \text{g}}{32.00 \, \text{g/mol}} = 0.1716 \, \text{mol} \]

Using the stoichiometry of the balanced equation, we compare the mole ratio of the reactants to find the limiting reactant.

• According to the balanced equation, 2 moles of \(\text{SO}_2\) react with 1 mole of \(\text{O}_2\).
• Required moles of \(\text{O}_2\) for 0.3933 moles of \(\text{SO}_2\): \[ \text{Required moles of O}_2 = \frac{0.3933 \, \text{mol SO}_2}{2} = 0.1967 \, \text{mol O}_2 \]

Since we only have 0.1716 moles of \(\text{O}_2\), \(\text{O}_2\) is the limiting reactant.

Using the limiting reactant to find the maximum mass of \(\text{SO}_3\) that can be formed.

• According to the balanced equation, 1 mole of \(\text{O}_2\) produces 2 moles of \(\text{SO}_3\).

• Moles of \(\text{SO}_3\) produced: \[ \text{Moles of SO}_3 = 2 \times 0.1716 \, \text{mol} = 0.3432 \, \text{mol} \]

• Molar mass of \(\text{SO}_3\): \[ \text{Molar mass of SO}_3 = 32.07 + 3 \times 16.00 = 80.07 \, \text{g/mol} \] \[ \text{Mass of SO}_3 = 0.3432 \, \text{mol} \times 80.07 \, \text{g/mol} = 27.47 \, \text{g} \]

Determine the amount of \(\text{SO}_2\) that reacts and the amount remaining.

• Moles of \(\text{SO}_2\) that react: \[ \text{Moles of SO}_2 = 2 \times 0.1716 \, \text{mol} = 0.3432 \, \text{mol} \]

• Initial moles of \(\text{SO}_2\): \[ 0.3933 \, \text{mol} \]

• Moles of \(\text{SO}_2\) remaining: \[ \text{Moles of SO}_2 \text{ remaining} = 0.3933 \, \text{mol} - 0.3432 \, \text{mol} = 0.0501 \, \text{mol} \]

• Mass of \(\text{SO}_2\) remaining: \[ \text{Mass of SO}_2 \text{ remaining} = 0.0501 \, \text{mol} \times 64.07 \, \text{g/mol} = 3.2095 \, \text{g} \]

Final Answer