Question

2 NaOH + H2SO4 ------> 2 H2O + Na2SO4

What mass of H2SO4 is required to react completely with 115 g NaOH?
Question 6 options:

141.0 g H2SO4

90 g H2SO4

180 g H2SO4

5 g H2SO4

Answer 1

To react completely with 115 g of NaOH, 141 g of H₂SO₄ is required. This is calculated by determining the moles of NaOH present and then using the stoichiometry of the balanced chemical equation to find the corresponding moles and mass of H₂SO₄ needed.

Step-by-step explanation:

To find the mass of H₂SO₄ required to react completely with 115 g of NaOH, we first need to use the balanced chemical equation: H₂SO₄ (aq) + 2NaOH (aq) → Na2SO4 (aq) + 2H₂O(l). According to the equation, 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide. The molar mass of NaOH is 40.00 g/mol, so we can calculate the moles of NaOH in 115 g as follows:

Moles of NaOH = mass of NaOH / molar mass of NaOH = 115 g / 40.00 g/mol = 2.875 moles NaOH.

From the balanced equation, we know the mole ratio of H₂SO₄ to NaOH is 1:2, so 2.875 moles of NaOH would require half the amount of H₂SO₄, which is 2.875 moles NaOH / 2 = 1.4375 moles H₂SO₄. The molar mass of H₂SO₄ is approximately 98.079 g/mol. Therefore, the mass of H₂SO₄ needed is:

Mass of H₂SO₄ = moles of H₂SO₄ * molar mass of H₂SO₄ = 1.4375 moles * 98.079 g/mol = 141 g H₂SO₄.

Answer 2

141.0g of H2SO4

Step-by-step explanation:

2 mole of NaOH reacts with 1 mole of H2SO4 completely

NaOH = 23 + 16 + 1 = 40g/mol

2 mole = 40g/mol x 2 mol = 80g

H2SO4 = 2(1) + 32 + 4(16) = 98g/mol

1 mole = 98g/mol x 1 mol = 98g

80g reacts with 98g completely

80g ------------ 98g

115g ------------ x

80x = 115 x 98

80x = 11270

x = 11270/80

x = 140.875g

x is approximately 141g

Therefore 115g of NaOH will react completely with 141g of H2SO4