Using 2NaOH + H2SO4 = 2H2O + Na2SO4

How many grams of sodium sulfate will be formed if you start with 150 grams of sodium hydroxide and you have an excess of sulfuric acid?

Respuesta :

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Answer:

266.25 grams of sodium sulfate will be formed if you start with 150 grams of sodium hydroxide and you have an excess of sulfuric acid.

Explanation:

2 NaOH + H₂SO₄ ⇒ 2 H₂O + Na₂SO₄

Observing the stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), you will see the amount of moles of reactant that react and product that is formed:

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

In the periodic table you can see the atomic mass of the elements that make up the compounds present in the reagents and products:

  • Na: 23 g/mole
  • O: 16 g/mole
  • H: 1 g/mole
  • S: 32 g/mole

Then, the molar mass of the compounds present in the reaction will be:

  • NaOH: 23 g/mole + 16 g/mole + 1 g/mole= 40 g/mole
  • H₂SO₄: 2*1 g/mole + 32 g/mole + 4*16 g/mole= 98 g/mole
  • H₂O: 2*1 g/mole+ 16 g/mole= 18 g/mole
  • Na₂SO₄: 2*23 g/mole + 32 g/mole + 4*16 g/mole= 142 g/mole

So, taking into account the stoichiometry of the reaction in moles, it is possible to say that mass stoichiometry is:

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

To determine the mass of sodium sulfate that will be formed if you start with 150 grams of sodium hydroxide, a rule of three is applied taking into account the stoichiometry of the reaction. The rule of three raised is: if by stoichiometry 80 g of sodium hydroxide form 142 g of sodium sulfate, 150 g of sodium hydroxide how many grams of sodium sulfate will they form?

[tex]mass of sodium sulfate=\frac{150 g*142 g}{80 g}[/tex]

mass of sodium sulfate= 266.25 grams

266.25 grams of sodium sulfate will be formed if you start with 150 grams of sodium hydroxide and you have an excess of sulfuric acid.