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Aluminum reacts with sulfuric acid, which is the automobile batteries. if 20.0 grams of Al is placed into a solution containing 115 grams of H2SO4, how many grams of hydrogen gas could be produced?

2007-07-19 05:17:06 · 5 answers · asked by Anonymous in Science & Mathematics Chemistry

5 answers

2Al + 3H2SO4 ==> Al2(SO4)3 + 3H2

=> 3 mols of H2 is produced from 2 mols of Al
=> 3 * 2.016 gm of H2 is produced from 2 * 26.982 gm Al
=> from 20 gm of Al , 3 * 2.016 * 20 / 2 * 26.982 gm H2 will be evolved

= 2.241 gm of H2 will be evolved.

2007-07-19 05:25:56 · answer #1 · answered by s0u1 reaver 5 · 1 0

Just FYI -- automotive batteries do NOT use aluminum. They are lead-acid batteries, because that reaction is reversible. The Aluminum-sulfuric acid reaction is **not** reversible, and is too vigorous to be safe in an auto.

But the others are correct -- the reaction you are looking for is:

2 Al + 3 H2SO4 == Al2(SO4)3 + 3 H2

So figure out how many moles of AL and how many moles of H2SO4 you have, determine the limiting reagent, and then figure out how much hydrogen is produced. Easy.

2007-07-19 06:23:58 · answer #2 · answered by Dave_Stark 7 · 0 0

All problems of this sort are solved the same way: construct the balanced equation, determine the molecular weights, determine the number of moles of the reactants, determine the limiting reactant, and figure out how much product is produced in moles, and then in grams. I will give you the equation:
2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2.

2007-07-19 05:24:30 · answer #3 · answered by Anonymous · 1 0

2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2

20 g will take up 20*3*98/54 = 33.06 g of H2SO4

So, gms of H2 = 20*3*2/54 = 2.22 g

2007-07-19 05:30:47 · answer #4 · answered by ag_iitkgp 7 · 1 0

The molecular weight of H2SO4 is ninety 8.a million g/mol. 250g (a million g/ninety 8.a million mol) = 2.fifty 5 mol H2SO4 The ratio of H2SO4:Al(SO4)3 is 3:a million, for this reason divide 2.fifty 5/3 for the mols created from Al(SO4)3. we've adequate for 0.eighty 5 mols aluminium sulfate. The molecular weight of Al(SO4)3 is 342.2 g/mol, so multiply (342.2)(0.eighty 5) That equals 290.7 grams of Al(SO4)3. solid success in Chemistry :)

2016-10-22 01:34:47 · answer #5 · answered by Anonymous · 0 0

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