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How many grams of sodium sulfate will be formed if you start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid?

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PhyCS
The entire process, assuming their degrees of ionization to be 1 due to their strong acid/base properties, should be

2NaOH + H2SO4 --> 2H2O + Na2SO4

Therefore, if 200 grams of sodium hydroxide are used with molar mass 40,  than 5 mols of sodium hydroxide was used. Using the ratio of 2:1:2:1, we see that 2.5 mols of Na2SO4 was used, therefore 2.5*molar mass = 2.5*142 = 355 grams.
CintiaSalazar

Taking into account the reaction stoichiometry, 355 grams of Na₂SO₄ are formed when 200 grams of sodium hydroxide reacts with excess of 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 Na₂SO₄ formed

The following rule of three can be applied: if by reaction stoichiometry 80 grams of NaOH form 142 grams of Na₂SO₄, 200 grams of NaOH form how much mass of Na₂SO₄?

[tex]mass of Na_{2} SO_{4} =\frac{200 grams of NaOHx142 grams of Na_{2} SO_{4}}{80 grams of NaOH}[/tex]

mass of Na₂SO₄= 355 grams

Then, 355 grams of Na₂SO₄ are formed when 200 grams of sodium hydroxide reacts with excess of H₂SO₄.

Learn more about the reaction stoichiometry:

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