Final answer:
The mass of Ag2SO4 formed when AgNO3 reacts with H2SO4 is calculated based on the moles of the limiting reactant. However, the calculated mass of 31.81 g does not match the provided multiple-choice options, indicating a possible error in the question.
Step-by-step explanation:
To determine the mass of Ag2SO4 that can be formed when AgNO3 reacts with H2SO4, we first need to write the balanced chemical equation for the reaction:
2 AgNO3(aq) + H2SO4(aq) → Ag2SO4(s) + 2 HNO3(aq)
We then calculate the moles of each reactant present to identify the limiting reactant. The molar mass of AgNO3 is approximately 169.87 g/mol, which gives:
Moles of AgNO3 = mass (g) / molar mass (g/mol) = 34.7 g / 169.87 g/mol = 0.204 moles
Given the stoichiometry of the balanced equation, we need twice this amount of AgNO3 to react with H2SO4 indicating AgNO3 is the limiting reactant. Now, we'll calculate the mass of Ag2SO4 formed using its molar mass (311.8 g/mol):
Mass of Ag2SO4 = moles of Ag2SO4 × molar mass (g/mol) = 0.102 moles × 311.8 g/mol = 31.81 grams
Since this is not one of the choices provided, we must carefully re-evaluate the stoichiometry and calculations. If any of the choices were close, we would round to the nearest option. Given the choices offered, it seems there may have been a typo or error in the question or choices provided.