When solutions of H₂SO₄ and NaOH react, the balanced molecular equation is: H₂SO₄(aq) + 2 NaOH(aq) → Na₂SO₄(aq) + 2 H₂O(l)
How much Na₂SO₄ is obtained when 4.00 g of H₂SO₄ reacts with 2.00 g of NaOH?

Question

Quizplus · Accepted Answer

To solve this problem, we need to use stoichiometry, which involves using the balanced equation to relate the amounts of reactants and products. First, we need to determine which reactant is limiting, which means it will be completely used up in the reaction. To do this, we will use the mole ratio between H₂SO₄ and NaOH from the balanced equation: 1 mol H₂SO₄ : 2 mol NaOH To convert the given masses of each substance into moles, we will use their respective molar masses: H₂SO₄: 96.10 g/mol NaOH: 40.00 g/mol 4.00 g H₂SO₄ x (1 mol / 96.10 g) = 0.0416 mol H₂SO₄ 2.00 g NaOH x (1 mol / 40.00 g) = 0.0500 mol NaOH Since the mole ratio is 1:2, it is clear that NaOH is the limiting reactant, since there are fewer moles of NaOH than H₂SO₄. Now that we know NaOH is limiting, we can use its moles to calculate how many moles of Na₂SO₄ are produced. From the balanced equation, we can see that the mole ratio between NaOH and Na₂SO₄ is also 1:1. 0.0500 mol NaOH x (1 mol Na₂SO₄ / 2 mol NaOH) = 0.0250 mol Na₂SO₄ Finally, we can convert the moles of Na₂SO₄ to grams: 0.0250 mol Na₂SO₄ x (128.18 g / 1 mol) = 3.205 g Na₂SO₄ Therefore, the answer is (C) 3.55 g. However, this is not one of the given choices, so the closest answer is (A) 0.650 g, which is the result if we had assumed that H₂SO₄ was the limiting reactant.

When Solutions of H2SO4 and NaOH React, the Balanced Molecular

When Solutions of H₂SO₄ and NaOH React, the Balanced Molecular