In the following equations, determined which reactant is the limiting reactant and which reactant is in excess. The amount used are given below. 2NaOH+H2SO4->NaSO4+2HO 15.4g NaOH and 15.4 H2SO4 react.
To determine the limiting reactant, we need to compare the amount of reactants used to the stoichiometry of the balanced equation.
Given:
- 15.4g of NaOH
- 15.4g of H2SO4
Using the molar mass of each compound:
- Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
- Molar mass of H2SO4 = 1.01 + 32.07 + (4 × 16.00) = 98.09 g/mol
Now, we can calculate the number of moles for each reactant:
- Moles of NaOH = 15.4g / 40.00 g/mol = 0.385 mol
- Moles of H2SO4 = 15.4g / 98.09 g/mol = 0.157 mol
Based on the balanced equation:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
We can see that the stoichiometry of the reaction is 2:1 for NaOH and H2SO4. This means that for every 2 moles of NaOH, we need 1 mole of H2SO4.
Comparing the number of moles available:
- Moles of NaOH: 0.385 mol
- Moles of H2SO4: 0.157 mol
Since the ratio required is 2:1, it means that we need 2 * 0.157 mol = 0.314 mol of NaOH to react with all the H2SO4. However, we only have 0.385 mol of NaOH available. This means that NaOH is the excess reactant.
Therefore, NaOH is in excess, and H2SO4 is the limiting reactant.