What is the minimum amount of 6.6 M H2SO4 necessary to produce 22.6 g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
I need to find the volume of the H2SO4
Work:
22.6g H2 /2.008g H2 = 11.3 mol H2
11.3 mol H2 * (3molH2SO4/3molH2) * (6.6molH2SO4/1L)= 74.58 mol H2SO4
From here I am lost
You are ok up to 11.3 mols H2 and 11.3 mols H2SO4. Then M = mols/L or L = mols/M = 11.3/6.6 = ? L
But how does one compounds mol is capable of being used with an entirely different compound of mols?
I don't know what you are talking about and you know your really are confused. The Al reacts with H2SO4 to produce hydrogen gas. All you are doing is saying that if you have 11.3 mols H2 gas produced it must have used 11.3 mols H2SO4 (which you calculated quite easily and properly). So you now want to convert mols H2SO4 to volume if you have 6.6 M to begin with. You are using stoiochiometry and the equation to convert mols H2 to mols H2SO4 (that's the mols H2 to mols H2SO4) and once you have mols H2SO4, you know M of anything = mols anything/L anything. You know mols and M, solve for L. You should be confused about the difference; in fact you calculated that part yourself.
What I was confused on was how I could use the mols of H2 and divide it by the mols of H2SO4 to find the volume. But after reading what you wrote, I understand now. Thanks!
To find the volume of the 6.6 M H2SO4 solution required, you can use the molarity formula:
Molarity (M) = moles of solute / volume of solution (in liters)
In this case, you have already calculated the number of moles of H2SO4 required (74.58 mol H2SO4). Now you need to rearrange the molarity formula to solve for volume:
Volume (in liters) = moles of solute / molarity
Substituting the values:
Volume = 74.58 mol H2SO4 / 6.6 M
Calculating the volume:
Volume = 11.33 L
Therefore, you need a minimum volume of 11.33 liters of 6.6 M H2SO4 to produce 22.6 g of H2 gas.