4. Part of the SO_(2) that is introduced into the atmosphere by combustion of sulfur containing compounds endl up being converted to sulfuric acid, H_(2)SO_(4) underline ( )SO_(2)+underline ( )O_(2)+underline ( )H_(2)Oarrow underline ( )H_(2)SO_(4) a) Determine the limiting reactant given that 250 mol SO_(2) is reacting with 24.0 mol O_(2) and 40.0 mol H_(2)O Using this information, what is the mass of H_(2)SO_(4) that can be formed? Answer: __ b) If 86gof H_(2)SO_(4) is produced, what were the amounts of each of the reactants? Answer: __

a) To determine the limiting reactant, we need to compare the mole ratio of the reactants with the stoichiometry of the balanced chemical equation.<br /><br />The balanced chemical equation for the conversion of $SO_{2}$ to $H_{2}SO_{4}$ is:<br /><br />$2SO_{2} + O_{2} + 2H_{2}O \rightarrow 2H_{2}SO_{4}$<br /><br />Given that 250 mol $SO_{2}$ is reacting with 24.0 mol $O_{2}$ and 40.0 mol $H_{2}O$, we can calculate the mole ratio of the reactants:<br /><br />Mole ratio of $SO_{2}$ to $O_{2}$: $\frac{250}{24.0} = 10.42$<br />Mole ratio of $SO_{2}$ to $H_{2}O$: $\frac{250}{40.0} = 6.25$<br /><br />Comparing the mole ratios with the stoichiometry of the balanced equation, we can see that the mole ratio of $SO_{2}$ to $O_{2}$ is greater than the stoichiometry, while the mole ratio of $SO_{2}$ to $H_{2}O$ is less than the stoichiometry. Therefore, $H_{2}O$ is the limiting reactant.<br /><br />To calculate the mass of $H_{2}SO_{4}$ that can be formed, we can use the stoichiometry of the balanced equation:<br /><br />$2SO_{2} + O_{2} + 2H_{2}O \rightarrow 2H_{2}SO_{4}$<br /><br />From the balanced equation, we can see that 2 mol $SO_{2}$ reacts with 1 mol $O_{2}$ and 2 mol $H_{2}O$ to produce 2 mol $H_{2}SO_{4}$. Since $H_{2}O$ is the limiting reactant, we can use its amount to calculate the amount of $H_{2}SO_{4}$ formed:<br /><br />$40.0 \text{ mol } H_{2}O \times \frac{2 \text{ mol } H_{2}SO_{4}}{2 \text{ mol } H_{2}O} = 40.0 \text{ mol } H_{2}SO_{4}$<br /><br />The molar mass of $H_{2}SO_{4}$ is 98.09 g/mol, so the mass of $H_{2}SO_{4}$ that can be formed is:<br /><br />$40.0 \text{ mol } H_{2}SO_{4} \times 98.09 \text{ g/mol } = 3923.6 \text{ g } H_{2}SO_{4}$<br /><br />Therefore, the mass of $H_{2}SO_{4}$ that can be formed is 3923.6 g.<br /><br />b) If 86 g of $H_{2}SO_{4}$ is produced, we can use the stoichiometry of the balanced equation to calculate the amounts of each of the reactants.<br /><br />From the balanced equation, we can see that 2 mol $SO_{2}$ reacts with 1 mol $O_{2}$ and 2 mol $H_{2}O$ to produce 2 mol $H_{2}SO_{4}$. The molar mass of $H_{2}SO_{4}$ is 98.09 g/mol, so 86 g of $H_{2}SO_{4}$ corresponds to:<br /><br />$\frac{86 \text{ g } H_{2}SO_{4}}{98.09 \text{ g/mol }} = 0.876 \text{ mol } H_{2}SO_{4}$<br /><br />Since the stoichiometry of the balanced equation is 2 mol $SO_{2}$ for every 2 mol $H_{2}SO_{4}$, the amount of $SO_{2}$ required is:<br /><br />$0.876 \text{ mol } H_{2}SO_{4} \times \frac{2 \text{ mol } SO_{2}}{2 \text{ mol } H_{2}SO_{4}} = 0.876 \text{ mol } SO_{2}$<br /><br />Similarly, the amount of $O_{2}$ required is:<br /><br />$0.876 \text{ mol } H_{2}SO_{4} \times \frac{1 \text{ mol } O_{2}}{2 \text{ mol } H_{2}SO_{4}} = 0.438 \text{ mol } O_{2}$<br /><br />Therefore, the amounts of each of the reactants required to produce 86 g of $H_{2}SO_{4}$ are 0.876 mol $SO_{2}$ and 0.438 mol $O_{2}$.