Elemental S reacts with O2 to form SO3 according t...

Question

Elemental S reacts with O2 to form SO3 according to the reaction2 S + 3 O2 → 2 SO3A) How many O2 molecules are needed to react with 4.13 g of S? Express your answer numerically in units of molecules.B) What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.C) Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up. For each of the given situations, indicate whether S or O2 is the limiting reactant.• 3.00 mol Sulfur, 3.00 mol Oxygen• 3.00 mol Sulfur, 4.00 mol Oxygen• 3.00 mol Sulfur, 5.00 mol Oxygen

Jules Bruno · Accepted Answer

We’re given the following balanced reaction:2 S + 3 O2 → 2 SO3Part A: We’re being asked to calculate the O2 molecules needed to react with 4.13 g of S.We need to perform a mole-to-mole comparison between O2 and S. molar mass of S = 32.066 g/mol1 mole = 6.022x1023 molecules[readmore]4.13 g S×1 mol S32.066 g S×3 mol O22 mol S×6.022×1023 mol O21 mol O2= 1.163x1023 O2 molecules1.163x1023 O2 molecules are needed to react with 4.13 g S.For Part B: We're being asked to determine the theoretical yield of SO3 produced. Since we calculated the exact amount of O2 that can react with S, we can use either to calculate the theoretical yield. For this calculation we will use 4.13 g.molar mass of S = 32.066 g/molmolar mass of SO3 = 80.066 g/mol mass SO3=4.13 g S×1 mol S32.066 g S×2 mol SO32 mol S×80.066 g SO31 mol SO3mass SO3 = 10.31 g SO3The theoretical yield of SO3 produced is 10.31 g.For Part C: Notice that we are given the massoles of both reactants: this means we need to determine the limiting reactant, which is the reactant that forms the less amount of product. This is because once the limiting reactant is all used up, the reaction can no longer proceed and make more products.This means the limiting reactant determines the maximum mass of the product formed.We need to perform a mole-to-mole comparison between each reactant and SO3. • 3.00 mol Sulfur, 3.00 mol OxygenFor S: 3.00 mol S×2 mol SO32 mol S = 3.00 mol SO3For O2:3.00 mol O2×2 mol SO33 mol O2 = 2.00 mol SO3Since O2 forms fewer moles of SO3 than S, O2 is the limiting reactant.• 3.00 mol Sulfur, 4.00 mol OxygenFor S: 3.00 mol S×2 mol SO32 mol S = 3.00 mol SO3For O2:4.00 mol O2×2 mol SO33 mol O2 = 2.66 mol SO3Since O2 forms fewer moles of SO3 than S, O2 is the limiting reactant.• 3.00 mol Sulfur, 5.00 mol OxygenFor S: 3.00 mol S×2 mol SO32 mol S = 3.00 mol SO3For O2:5.00 mol O2×2 mol SO33 mol O2 = 3.33 mol SO3Since S forms fewer moles of SO3 than O2, S is the limiting reactant.

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