320g of sulfur dioxide react with 32.0g of oxygen and excess water to form sulfuric acid (H2SO4)
a. What is the limiting reagent?
b. What is the theoretical yield of product?
c. How many grams of excess reagent will remain?
2SO2+ O2 + 2H2O >>> 2H2SO4
320 gram SO2=5 moles check that
32 g O2= one mole
So looking at the balanced equation, you have an excess of SO2, or a deficit of O2.
To determine the limiting reagent, we need to compare the moles of each reactant and their stoichiometric ratios with the desired product, sulfuric acid (H2SO4).
Step 1: Calculate the moles of each reactant:
Moles of SO2 = mass of SO2 / molar mass of SO2
Moles of SO2 = 320g / 64.06g/mol = 5.00 mol
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 32.0g / 32.00g/mol = 1.00 mol
Step 2: Determine the stoichiometric ratio between the reactants and the product:
From the balanced chemical equation, we see that:
2 moles of SO2 reacts with 1 mole of O2 to produce 2 moles of H2SO4.
Step 3: Compare the moles of each reactant with the stoichiometric ratio:
For SO2:
Moles of H2SO4 that could be produced from SO2 = 5.00 mol SO2 * (2 mol H2SO4 / 2 mol SO2) = 5.00 mol H2SO4
For O2:
Moles of H2SO4 that could be produced from O2 = 1.00 mol O2 * (2 mol H2SO4 / 1 mol O2) = 2.00 mol H2SO4
a. The limiting reagent is the reactant that produces less moles of the desired product. In this case, O2 produces only 2.00 mol of H2SO4, while SO2 produces 5.00 mol of H2SO4. Therefore, O2 is the limiting reagent.
b. The theoretical yield of the product is the maximum amount of product that can be formed from the limiting reagent. Since O2 is the limiting reagent and it produces 2.00 mol of H2SO4, we can calculate the theoretical yield using the stoichiometric ratio:
Theoretical yield of H2SO4 = Moles of limiting reagent * (2 mol H2SO4 / 1 mol O2)
Theoretical yield of H2SO4 = 1.00 mol O2 * (2 mol H2SO4 / 1 mol O2) = 2.00 mol H2SO4
c. To calculate the grams of excess reagent remaining, we need to first determine the moles of the excess reagent that were not used in the reaction. In this case, SO2 is the excess reagent.
Moles of excess SO2 = Moles of initial SO2 - moles of SO2 consumed by the reaction
Moles of SO2 consumed by the reaction = Moles of limiting reagent * (2 mol SO2 / 1 mol H2SO4)
Moles of SO2 consumed by the reaction = 1.00 mol O2 * (2 mol SO2 / 1 mol O2) = 2.00 mol SO2
Moles of excess SO2 = 5.00 mol SO2 - 2.00 mol SO2 = 3.00 mol SO2
Grams of excess SO2 remaining = Moles of excess SO2 * molar mass of SO2
Grams of excess SO2 remaining = 3.00 mol * 64.06 g/mol = 192.18 g of SO2 will remain.
To determine the limiting reagent, theoretical yield of product, and the grams of excess reagent remaining, we need to follow these steps:
1. Write the balanced chemical equation for the reaction:
SO2 + O2 + H2O -> H2SO4
2. Calculate the molar mass of each substance:
Molar mass of SO2 = 32.07 g/mol
Molar mass of O2 = 32.00 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of H2SO4 = 98.09 g/mol
3. Convert the mass of each reactant to moles:
Moles of SO2 = 320g / 32.07 g/mol = 9.99 mol
Moles of O2 = 32.0g / 32.00 g/mol = 1.00 mol
4. Determine the stoichiometric ratio (mole ratio) of the reactants based on the balanced equation:
From the equation, the ratio of SO2 to O2 is 1:1.
5. Identify the limiting reagent:
The limiting reagent is the reactant that is completely consumed first during the reaction. To determine this, compare the mole ratios of the reactants. In this case, the mole ratio of SO2 and O2 is 1:1. Since the mole ratios are the same, both SO2 and O2 will react completely without any excess remaining.
6. Calculate the theoretical yield of the product:
The theoretical yield is the maximum amount of product that can be formed based on the stoichiometry of the balanced equation. In this case, the stoichiometry tells us that the mole ratio of SO2 to H2SO4 is also 1:1. Therefore, the theoretical yield of H2SO4 is equal to the moles of SO2, which is 9.99 mol. To convert this to grams, multiply by the molar mass of H2SO4:
Theoretical yield of H2SO4 = 9.99 mol * 98.09 g/mol = 980.099 g or 980.1 g (rounded to one decimal place).
7. Calculate the grams of the excess reagent remaining:
Since there is no excess of either SO2 or O2, there will be no grams of excess reagent remaining.
Summary:
a. The limiting reagent is both SO2 and O2 since they are present in an equal molar ratio.
b. The theoretical yield of H2SO4 is 980.1 g.
c. There is no excess reagent remaining for either SO2 or O2.