How many grams of H2SO4 are needed to react with 78.86 grams of AI(NO3)3 ?
I

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?

[tex]mass of H_{2} SO_{4} =\frac{78.86 grams of Al(NO_{3} )_{3}*294gramsofH_{2} SO_{4} }{436 grams of Al(NO_{3} )_{3}}[/tex]

mass of H₂SO₄=53.18 g

53.18 grams of H₂SO₄ are needed to react with 78.86 grams of Al(NO₃)₃.

How many grams of H2SO4 are needed to react with 78.86 grams of AI(NO3)3 ? I

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How many grams of H2SO4 are needed to react with 78.86 grams of AI(NO3)3 ?
I