Final answer:
To react with 97.25 g of HgS, 0.418 moles of CaO are required based on stoichiometry and the balanced chemical equation.
Step-by-step explanation:
To determine how many moles of CaO are required to react with 97.25 g of HgS, we need to use stoichiometry based on the balanced chemical equation provided. Using the molecular weights of HgS and CaO, we can convert grams to moles and then use the stoichiometric ratios to find the answer.
First, we need to calculate the molar mass of HgS (mercury(II) sulfide), which is approximately 232.66 g/mol (200.59 for Hg + 32.07 for S). By dividing the given mass of HgS (97.25 g) by its molar mass, we can find out how many moles of HgS we have:
97.25 g HgS x (1 mol HgS / 232.66 g HgS) = 0.418 moles HgS
The balanced equation shows that we need 1 mole of CaO for every mole of HgS. So we require the same number of moles of CaO as HgS to complete the reaction:
0.418 moles HgS x (1 mol CaO / 1 mol HgS) = 0.418 moles CaO
Therefore, 0.418 moles of CaO are required to react with 97.25 g of HgS, which is the stoichiometric amount.