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?
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