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1) State the balanced chemical reaction

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

2) State the molar ratios:

1 mol H2SO4 : 2 mol NaOH

3) Set the proportions using the number of moles of one rectant and x for the unknown one>

1 mol H2SO4 / 2 mol NaOH = x mol H2SO4 / 0.75 mol NaOH

=> x = 0.75 mol NaOH * [1 mol H2SO4 / 2 mol NaOH]

x = 0.375 mol H2SO4

4) Calculate the mass using the molar mass of H2SO4, MM.

MM = 2*1g/mol + 32.1 g/mol + 4*16 g/mol = 98.1 g/mol

mass = # of moles * molar mass = 0.375 mol * 98.1 g/mol = 36.79 g.

Answer: 36.8 grams

 
 

 
CintiaSalazar

Taking into account the reaction stoichiometry:

  • 36,75 grams of H₂SO₄ is required to react with 0.75 mol of NaOH.
  • 53.25 grams of Na₂SO₄ and 13.5 grams of H₂O are formed when 0.75 moles of NaOH reacts with H₂SO₄.

Reaction stoichiometry

In first place, the balanced reaction is:

H₂SO₄ + 2 NaOH  → Na₂SO₄ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • H₂SO₄: 1 mole
  • NaOH: 2 moles  
  • Na₂SO₄: 1 mole
  • H₂O: 2 moles

The molar mass of the compounds is:

  • H₂SO₄: 98 g/mole
  • NaOH: 40 g/mole
  • Na₂SO₄: 142 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • H₂SO₄: 1 mole× 98 g/mole= 98 grams
  • NaOH: 2 moles× 40 g/mole= 80 grams
  • Na₂SO₄: 1 mole× 142 g/mole= 142 grams
  • H₂O: 2 moles× 18 g/mole= 36 grams

Mass of H₂SO₄ required

The following rule of three can be applied: If by reaction stoichiometry 2 moles of NaOH react with 98 grams of H₂SO₄, 0.75 moles of NaOH react with how much mass of H₂SO₄?

[tex]mass of H_{2}S O_{4} =\frac{0.75 moles of NaOHx98 grams of H_{2}S O_{4} }{2 moles of NaOH}[/tex]

mass of H₂SO₄= 36.75 grams

Finally, 36,75 grams of H₂SO₄ is required to react with 0.75 mol of NaOH.

Mass of each product formed

The following rules of three can be applied:

  • if by reaction stoichiometry 2 moles of NaOH form 142 grams of Na₂SO₄, 0.75 moles of NaOH form how much mass of Na₂SO₄?

[tex]mass of Na_{2}S O_{4} =\frac{0.75 moles of NaOHx142 grams of Na_{2}S O_{4} }{2 moles of NaOH}[/tex]

mass of Na₂SO₄= 53.25 grams

  • if by reaction stoichiometry 2 moles of NaOH form 36 grams of H₂O, 0.75 moles of NaOH form how much mass of H₂O?

[tex]mass of H_{2}O =\frac{0.75 moles of NaOHx36 grams of H_{2}O}{2 moles of NaOH}[/tex]

mass of H₂SO₄= 13.5 grams

Then, 53.25 grams of Na₂SO₄ and 13.5 grams of H₂O are formed when 0.75 moles of NaOH reacts with H₂SO₄.

Learn more about the reaction stoichiometry:

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