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Based on the equation below, how many moles of aluminum sulfate (Al2(SO4)3) will be produced from the reaction of 5.8 moles of aluminum hydroxide (Al(OH)3) and excess sulfuric acid (H2SO4)? 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Sagot :

CintiaSalazaridn

Answer:

2.9 moles of aluminum sulfate will be produced from the reaction of 5.8 moles of aluminum hydroxide and excess sulfuric acid.

Explanation:

The balanced reaction is:

2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O

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

  • Al(OH)₃: 2 moles
  • H₂SO₄: 3 moles
  • Al₂(SO₄)₃: 1 mole
  • H₂O: 6 moles

Then you can apply the following rule of three: if by reaction stoichiometry 2 moles of Al(OH)₃ produce 1 mole of Al₂(SO₄)₃, then 5.8 moles of Al(OH)₃ how many moles of Al₂(SO₄)₃ will they produce?

[tex]moles of Al_{2} (SO_{4} )_{3} =\frac{5.8moles of Al(OH)_{3} *1 mole of Al_{2} (SO_{4} )_{3}}{2moles of Al(OH)_{3}}[/tex]

moles of Al₂(SO₄)₃= 2.9

2.9 moles of aluminum sulfate will be produced from the reaction of 5.8 moles of aluminum hydroxide and excess sulfuric acid.

Answer:

67

Explanation:

because that right

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