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S8 + 12O2 -------> 8SO3

If you start with 873.2 g of S8 and 859.3 g of O2, what mass of SO3 will be produced?

User Karask
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To determine the mass of SO3 produced, we need to use stoichiometry to find the limiting reactant and the theoretical yield of SO3.

The balanced chemical equation is:

S8 + 12O2 → 8SO3

The molar mass of S8 is 256.52 g/mol. Therefore, the number of moles of S8 is:

n(S8) = mass / molar mass = 873.2 g / 256.52 g/mol = 3.4 mol

The molar mass of O2 is 32.00 g/mol. Therefore, the number of moles of O2 is:

n(O2) = mass / molar mass = 859.3 g / 32.00 g/mol = 26.85 mol

According to the balanced chemical equation, the stoichiometric ratio of S8 to O2 is 1:12. Therefore, the number of moles of O2 needed to react with all of the S8 is:

n(O2) needed = n(S8) × 12 = 3.4 mol × 12 = 40.8 mol

Since the number of moles of O2 available (26.85 mol) is less than the number of moles of O2 needed (40.8 mol), O2 is the limiting reactant.

The theoretical yield of SO3 is based on the number of moles of O2 used. Since 12 moles of O2 are required to produce 8 moles of SO3, the number of moles of SO3 produced is:

n(SO3) = n(O2) × (8/12) = 26.85 mol × (8/12) = 17.90 mol

The molar mass of SO3 is 80.06 g/mol. Therefore, the mass of SO3 produced is:

mass(SO3) = n(SO3) × molar mass = 17.90 mol × 80.06 g/mol = 1433 g

Therefore, the mass of SO3 produced is 1433 g.
User Neven
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