Consider the unbalanced chemical equation \(\mathrm{S}+\mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (a) Balance the equation. (b) If you react \(4.80 \mathrm{~g}\) of sulfur with \(16.20 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\), how many grams of \(\mathrm{SO}_{2}\) can theoretically be produced?

We have the unbalanced chemical equation: \[\mathrm{S}+\mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2}+\mathrm{H}_{2} \mathrm{O}\] To balance the equation, we need to make sure that the number of atoms of each element is the same on both sides. We can balance the given equation as: \[\mathrm{S} + \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow \mathrm{SO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] Now, the number of atoms for each element is the same on both sides of the equation.

Before we can figure out how much SO2 can be produced, we must first identify the limiting reactant - the reactant that will run out first. We need to compare the amounts of S and H2SO4 given with their stoichiometry (mole ratio) in the balanced equation. First, let's convert the given masses to moles: Given: Mass of S = 4.80 g Molar mass of S = 32 g/mol Moles of S = mass / molar mass = 4.80 g / 32 g/mol = 0.15 mol Mass of H2SO4 = 16.20 g Molar mass of H2SO4 = 98 g/mol (2 + 32 + 4 × 16) Moles of H2SO4 = mass / molar mass = 16.20 g / 98 g/mol = 0.165 mol Now, let's find the mole ratio of S and H2SO4 in the balanced chemical equation: Mole ratio of S : H2SO4 in the balanced equation = 1 : 1 Next, let's divide the moles of each reactant by their stoichiometric coefficient: 0.15 mol S / 1 = 0.15 0.165 mol H2SO4 / 1 = 0.165 The smallest value corresponds to the limiting reactant. In this case, S is the limiting reactant.

To find the amount of SO2 produced, we first determine the mole ratio of S (limiting reactant) to SO2 in the balanced equation. Then, we use the moles of the limiting reactant to find the moles of SO2 formed: Mole ratio of S : SO2 in the balanced equation = 1 : 1 Moles of SO2 formed = moles of S (limiting reactant) = 0.15 mol Now, we can convert moles of SO2 to grams: Molar mass of SO2 = 64 g/mol (32 + 2 × 16) Mass of SO2 produced = moles × molar mass = 0.15 mol × 64 g/mol = 9.6 g Thus, 9.6 grams of SO2 can theoretically be produced.