How many moles of sulfuric acid are needed to produce 57.8 milliliters of water? Water has the density of 0.987 g/mL. Show all steps of your calculation as well as the final answer.

NaOH + H2SO4 → Na2SO4 + H2O

My answer:

Balance the equation: 2NaOH + H2SO4 = Na2SO4 + 2H2O

mass of water = 57.8 mL x 1 g/mL = 57.8g

moles of water = 57.8g / 18.01 g/mol = 3.21

The ratio between H2SO4 and H2O is 1:2

moles of H2SO4 required = 3.21/2 = 1.61

Question

Cheyenne Goodwin

Chemistry

27 September, 20:17

How many moles of sulfuric acid are needed to produce 57.8 milliliters of water? Water has the density of 0.987 g/mL. Show all steps of your calculation as well as the final answer.

NaOH + H2SO4 → Na2SO4 + H2O

My answer:

Balance the equation: 2NaOH + H2SO4 = Na2SO4 + 2H2O

mass of water = 57.8 mL x 1 g/mL = 57.8g

moles of water = 57.8g / 18.01 g/mol = 3.21

The ratio between H2SO4 and H2O is 1:2

moles of H2SO4 required = 3.21/2 = 1.61

+4

Answers (1)

Know the Answer?

Maya Blackburn · Answer 1

The balanced chemical reaction is:

2NaOH + H2SO4 → Na2SO4 + 2H2O

We first need to determine the amount of water to be produced by using the density and the molar mass water. Then, we use the relation from the reaction to relate water with sulfuric acid.

57.8 mL (0.987 g/mL) = 57.05 g H2O

57.05 g H2O (1 mol / 18.02 g) = 3.17 mol H2O

3.17 mol H2O (1 mol H2SO4 / 2 mol H2O) = 1.58 mol H2SO4 needed