Answer:
53.18 grams of H₂SO₄ are needed to react with 78.86 grams of Al(NO₃)₃.
Step-by-step explanation:
The balanced reaction between Al(NO₃)₃ and H₂SO₄ is:
2 Al(NO₃)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 HNO₃
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents are required:
- Al(NO₃)₃: 2 moles
- H₂SO₄: 3 moles
Being the molar mass of the elements:
- Al: 27 g/mole
- N: 14 g/mole
- O: 16 g/mole
- H: 1 g/mole
- S: 32 g/mole
then the molar mass of the reactants are:
- Al(NO₃)₃: 27 g/mole + 3*(14 g/mole + 3*16 g/mole)= 213 g/mole
- H₂SO₄: 2*1 g/mole + 32 g/mole +4*16 g/mole= 98 g/mole
Then, by reaction stoichiometry, the following reagent mass amounts are required:
- Al(NO₃)₃: 2 moles* 213 g/mole= 436 g
- H₂SO₄: 3 moles*98 g/mole= 294 g
Then you can apply the following rule of three: If by stoichiometry 436 g of Al(NO₃)₃ react with 294 g of H₂SO₄, 78.86 g of Al(NO₃)₃ with how much mass of H₂SO₄ will it react?
mass of H₂SO₄=53.18 g
53.18 grams of H₂SO₄ are needed to react with 78.86 grams of Al(NO₃)₃.