Reaction equation question?

Question

Aluminum reacts with H2CO3 to produce H2 and Al 2(CO3)3. How many grams of aluminum would be needed to produce 9.00 grams of H2 gas?

I need a step by step to understand if anyone can help.

Answer 1

Molar mass of H2 = 2 g/mol

Mol = 9 grams : 2g/mol = 18 mol

The reactions :

2Al + 3H2CO3 --> 3H2 + Al2(CO3)3

Mol of Al = 2/3 x 18 mol = 12 mol

Molar mass of Al = 27 g/mol

Mass of Al = 12 mol x 27 g/mol

= 324 grams

Answer 2

2AL + 3H2CO3 ------   Al2(CO3)3 + 3 H2

2 moles of Al will react with H2CO3 to produce 3 moles of H2

so 9 grams of H2/2 grams /mole = 4.5 moles of H2 gas

therefore if 2 moles Al/3 moles of H2 = x moles Al/4 moles of H2 gas..  2moles  X 4.5moles/3 moles =  3 moles ..

so 3 moles of Al are needed to produce 9 grams ( 4.5 moles of H2 )   3 moles X 27 grams/mole =  81 grams

Proof

54 grams of Al  yield 6 grams of H2  ( 2 moles Al yield 3 moles of H2 )  so

54/6 = X/9 grams = 81 grams