Aluminum reacts with sulfuric acid, which is the acid in car batteries. If 20.0 grams of Al is placed into a solution contain 115 gram of H2SO4, how many grams of hydrogen ga could be be produced?

2Al + 3H2SO4 ---> Al(SO4)3 + 3H2

20.0g/27 = .741
.741 X 3/2 = 1.11
1.11 X 2 = 2.22

im confused on what to do next

This is a limiting reagent type problem. First, you omitted a 2 from the Al2(SO4)3 but I expect that's just a typo.
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
20.0g/27 = .741
.741 X 3/2 = 1.11
1.11 X 2 = 2.22 I assume this is mols H2 x molar mass H2 to produce 2.22 grams H2. It looks good to here. Actually, this is the end of the problem, EXCEPT, you haven't determined if Al or H2SO4 is the limiting reagent.
I do that this way.
You have already determined that 1.11 mol H2 can be produced from 20 g Al. Now we need to determine the mols H2 that can be produced from 115 grams H2SO4. Its done the same say.
115g/98 =1.17 mols H2SO4
1.17 mols H2SO4 x (3 mols H2/3 mols H2SO4) = 1.17 x 1/1 = 1.17 mols H2.
Now compare the mols from Al and the mols from H2SO4. Both answers can't be correct, of course, and the SMALLER of the two is always the one that is the limiting reagent. Therefore, Al is the limiting reagent and some of the H2SO4 will remain unreacted so you are through with the problem. Had it been that H2SO4 was the limiting reagent, then we would have gone through, x'd out all of the Al work and substituted the mols and grams H2 from the H2SO4 data.


so the answer is .741?

no my mistake 1.11?

No, the answer is 2.22 grams H2.

No, the answer is not 0.741.

The correct answer is 2.22 grams of hydrogen gas (H2) could be produced. This is based on the stoichiometry of the balanced equation where 2 moles of aluminum (Al) react with 3 moles of sulfuric acid (H2SO4) to produce 3 moles of hydrogen gas (H2).

To calculate the grams of hydrogen gas produced, you need to convert the moles of hydrogen gas to grams using the molar mass of hydrogen (H2), which is 2.02 g/mol.

So, 1.11 moles of hydrogen gas x 2.02 g/mol = 2.24 grams of hydrogen gas.

Therefore, the correct answer is 2.24 grams of hydrogen gas could be produced.

No, the correct answer is 2.22 grams of hydrogen gas could be produced in this reaction. By calculating the stoichiometry of the reaction, you correctly determined that 1.11 moles of hydrogen gas can be produced from 20 grams of aluminum. However, in order to determine the limiting reagent and the actual amount of hydrogen gas produced, we need to compare this value to the amount of hydrogen gas that can be produced from the sulfuric acid.

To do this, we calculate the number of moles of hydrogen gas that can be produced from the given 115 grams of sulfuric acid:
115g/98g/mol = 1.17 moles of H2SO4
1.17 moles of H2SO4 x (3 moles of H2/3 moles of H2SO4) = 1.17 moles of H2

Since the number of moles of H2 from the aluminum (1.11 moles) is smaller than the number of moles of H2 from the sulfuric acid (1.17 moles), aluminum is the limiting reagent. This means that all 1.11 moles of H2 from the aluminum will be completely used up, and the amount of hydrogen gas produced is limited by the amount of aluminum.

To find the mass of hydrogen gas produced, we convert the moles of H2 to grams:
1.11 moles x 2g/mol = 2.22 grams of hydrogen gas

Therefore, 2.22 grams of hydrogen gas can be produced in this reaction.