What is the minimum amount of 5.0M H2SO4 necessary to produce 24.1g of H2(g) according to the following reaction? 2Al(s) + 3H2SO4(aq) produces Al2(SO4)3(aq)+3H2(g).
Express your answer using two significant figures.
I will be happy to critique your thinking. figure out how many moles of H2 are produced first. Then, use the fact that 3 moles H2SO4 produces 3 moles H2.
To find the minimum amount of 5.0M H2SO4 necessary to produce 24.1g of H2 gas, we need to use stoichiometry and the molarity of the H2SO4 solution.
First, let's calculate the moles of H2 gas produced using the given mass:
Mass of H2 = 24.1g
To convert grams to moles, we need to use the molar mass of H2, which is approximately 2.0 g/mol.
Moles of H2 = Mass of H2 / Molar mass of H2
= 24.1g / 2.0 g/mol
= 12.05 mol
According to the balanced equation, the ratio between H2SO4 and H2 is 3:1. That means for every 3 moles of H2SO4, 1 mole of H2 gas is produced.
So, the moles of H2SO4 needed can be calculated as follows:
Moles of H2SO4 = (3 moles of H2SO4 / 1 mole of H2) * (12.05 moles of H2)
= 36.15 moles
Now, we need to find the volume of the 5.0M H2SO4 solution that contains 36.15 moles of the acid. To do this, we use the equation:
Molarity = Moles / Volume
We rearrange this equation to solve for Volume:
Volume = Moles / Molarity
= 36.15 moles / 5.0 mol/L
= 7.23 L
Therefore, the minimum amount of 5.0M H2SO4 necessary to produce 24.1g of H2 gas is 7.23 liters, when considering the two significant figures in the given molarity value.