For the balanced equation shown below, if 40.1 grams of C2H3O2Br were reacted with 8.01 grams of O2, how many grams of H2O would be produced?
2C2H3O2Br+O2->4CO+2H2O+2HBr
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To solve this problem, we need to use the given balanced equation and calculate the stoichiometry of the reaction.
1. Start by calculating the number of moles for each reactant:
- Moles of C2H3O2Br = mass (g) / molar mass (g/mol)
- Moles of C2H3O2Br = 40.1 g / (12.01 g/mol + 2 * 1.01 g/mol + 2 * 16.00 g/mol + 79.90 g/mol) ≈ 0.228 mol
- Moles of O2 = mass (g) / molar mass (g/mol)
- Moles of O2 = 8.01 g / (2 * 16.00 g/mol) ≈ 0.251 mol
2. Determine the limiting reactant. This can be done by comparing the mole ratio between C2H3O2Br and O2 in the balanced equation:
From the balanced equation, we can see that the mole ratio of C2H3O2Br to O2 is 2:1. Therefore, 2 moles of C2H3O2Br react with 1 mole of O2.
As we calculated earlier, we have 0.228 mol of C2H3O2Br and 0.251 mol of O2. The mole ratio of C2H3O2Br to O2 tells us that we have more O2 than required.
Therefore, O2 is in excess, and C2H3O2Br is the limiting reactant.
3. Now, use the stoichiometric ratio from the balanced equation to find the number of moles of H2O produced:
From the balanced equation, we can see that the mole ratio of C2H3O2Br to H2O is 2:2, which can be simplified to 1:1.
Therefore, the number of moles of H2O produced is equal to the number of moles of C2H3O2Br reacted, which is 0.228 mol.
4. Finally, calculate the mass of H2O produced:
Mass of H2O = moles of H2O * molar mass of H2O
Mass of H2O = 0.228 mol * (2 * 1.01 g/mol + 16.00 g/mol) ≈ 9.29 g
Therefore, approximately 9.29 grams of H2O would be produced when 40.1 grams of C2H3O2Br are reacted with 8.01 grams of O2.