Question

What is the mininium amount of 6.0 M H2SO4 necessary to produce 25.0 g of H2 (g) according to the reaction between aluminum and sulfuric acid? 2 Al(s) + 3 H2SO4(aq) ----> AL2(SO4)3(aq) + 3 H2(g) I know the answer is 2.1L. I don't understand how they got that answer.

Answer 1

2 Al(s) + 3 H2SO4(aq) ----> AL2(SO4)3(aq) + 3 H2(g)

from this balanced equation it is clear that

3 moles of H2SO4 required to produce the 3 moles of H2 means

1 moles of H2SO4 required to produce the 1 mole of H2

in problem he said to produce 25 grams convert in to moles

no of moles of H2 = weight of H2 / molar mass of H2

= 25 / 2 = 12.5 moles of H2

from balanced equation to produce 12.5 moles of H2 we need 12.5 moles of H2SO4 right?

so no of moles of H2SO4 = 12.5 moles

now we know the moles of H2So4 = 12.5 moles and concentration he has given 6.0 M

now we need to find out the volume of H2So4 right?

Molarity = no of moles / volume in liters

Volume in liters = 12.5 moles / 6.0M

= 2.083 L means = 2.1L of 6.0M H2So4