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
Thank you!
To find the number of grams of H2O produced, we first need to determine the limiting reactant, which is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
1. Calculate the number of moles for each reactant:
- Moles of C2H3O2Br = mass (g) / molar mass (g/mol)
- Moles of O2 = mass (g) / molar mass (g/mol)
2. Use the coefficients of the balanced equation to determine the mole ratio between the limiting reactant and the product of interest (H2O). From the balanced equation, we can see that the stoichiometric ratio between C2H3O2Br and H2O is 2:2 (or 1:1).
3. Identify the limiting reactant:
- Compare the mole ratios of each reactant to the product using their coefficients.
- The reactant that produces the least amount of product is the limiting reactant.
4. Calculate the number of moles of H2O produced using the mole ratio obtained in step 2.
5. Convert the number of moles of H2O to grams by multiplying the moles by the molar mass of H2O.
Let's calculate it step by step!
1. Moles of C2H3O2Br:
Molar mass of C2H3O2Br = (2 * 12.01 g/mol) + (3 * 1.01 g/mol) + (1 * 79.90 g/mol) + (1 * 79.90 g/mol)
= 122.94 g/mol
Moles of C2H3O2Br = 40.1 g / 122.94 g/mol
2. Moles of O2:
Molar mass of O2 = (2 * 16.00 g/mol)
= 32.00 g/mol
Moles of O2 = 8.01 g / 32.00 g/mol
3. Limiting reactant:
To determine the limiting reactant, we need to compare the mole ratios.
From the balanced equation, the mole ratio of C2H3O2Br to H2O is 2:2.
Moles of H2O from C2H3O2Br = (moles of C2H3O2Br) * (2 moles H2O / 2 moles C2H3O2Br)
Moles of H2O from O2 = (moles of O2) * (2 moles H2O / 1 mole O2)
Compare the moles of H2O produced from each reactant:
- If the moles of H2O from C2H3O2Br are greater, then C2H3O2Br is the limiting reactant.
- If the moles of H2O from O2 are greater, then O2 is the limiting reactant.
4. Calculate the moles of H2O produced:
If C2H3O2Br is the limiting reactant, use the moles of H2O from C2H3O2Br.
If O2 is the limiting reactant, use the moles of H2O from O2.
5. Convert moles of H2O to grams:
Multiply the number of moles of H2O by the molar mass of H2O.
By following these steps, you can find the grams of H2O produced in this reaction.
Here is a solved simple stoichiometry problem. You have a limiting reagent problem. The way I solve these is to solve two simple stoichiometry problems, one for each of the reagent you are given. You will get two different amounts (moles) of product formed; the correct answer in limiting reagent problem is ALWAYS the smaller one and the reagent providing that value is the limiting reagent.
http://www.jiskha.com/science/chemistry/stoichiometry.html