What is the minimum amount of 6.3 M H2SO4 necessary to produce 27.7 g of H2 (g) according to the following reaction?
2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)
mols H2 needed = grams/molar mass =( 27.7/2) = 13.85
Convert mols H2 to mols H2SO4 this way:
13.35 mols H2 x (3 mols H2SO4/3 mols H2) = 13.35 mols H2SO4,
Then M H2SO4 = mols H2SO4/L H2SO4.
Yu know mols and M, solve for L. Convert to mL if needed.
To find the minimum amount of 6.3 M H2SO4 necessary to produce 27.7 g of H2 gas (g), we need to use stoichiometry.
First, let's write out the balanced chemical equation:
2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)
From the balanced equation, we can see that the stoichiometric ratio between H2SO4 and H2 is 3:3 (or 1:1). This means that for every 3 moles of H2SO4, we get 3 moles of H2.
Step 1: Calculate the moles of H2
The molar mass of H2 is 2 g/mol.
moles of H2 = mass / molar mass
moles of H2 = 27.7 g / 2 g/mol
moles of H2 ≈ 13.85 mol
Step 2: Calculate the moles of H2SO4
Since the stoichiometric ratio of H2SO4 to H2 is 1:1, the moles of H2SO4 will be the same as the moles of H2.
moles of H2SO4 = 13.85 mol
Step 3: Calculate the volume of 6.3 M H2SO4
Now we need to convert the moles of H2SO4 into volume using the molarity (M) and the formula:
moles of solute = M x volume of solution (in L)
volume of solution (in L) = moles of solute / M
volume of 6.3 M H2SO4 = 13.85 mol / 6.3 M
volume of 6.3 M H2SO4 ≈ 2.20 L
Therefore, the minimum amount of 6.3 M H2SO4 necessary to produce 27.7 g of H2 is approximately 2.20 liters.