Question

How many moles of which excess reactant are present when 365.0 mL of 0.226 M sulfuric acid reacts with 0.510 L of 0.164 M sodium hydroxide to form water and aqueous sodium sulfate?

Answer 1

To determine the number of moles of sulfuric acid and sodium hydroxide, we need to use the balanced chemical equation and the given concentrations and volumes of the reactants. The number of moles of sulfuric acid is 0.08279 moles and the number of moles of sodium hydroxide is 0.08364 moles.

Step-by-step explanation:

To determine the number of moles of sulfuric acid and sodium hydroxide, we need to use the balanced chemical equation and the given concentrations and volumes of the reactants.

The balanced chemical equation for the reaction is:

H₂SO₄(aq) + 2NaOH(aq) → Na₂SO₄(aq) + 2H₂O(l)

First, we will calculate the number of moles of sulfuric acid:

1. Convert the volume of sulfuric acid to liters by dividing by 1000: 365.0 mL = 365.0/1000 = 0.365 L
2. Calculate the number of moles using the formula: moles = concentration × volume
3. moles of H₂SO₄ = (0.226 M) × (0.365 L)
4. = 0.08279 moles

Next, we will calculate the number of moles of sodium hydroxide:

1. Convert the volume of sodium hydroxide to liters by dividing by 1000: 0.510 L
2. Calculate the number of moles using the formula: moles = concentration × volume
3. moles of NaOH = (0.164 M) × (0.510 L)
4. = 0.08364 moles

Therefore, there are 0.08279 moles of sulfuric acid and 0.08364 moles of sodium hydroxide present.