What is the minimum amount of 5.9M H2SO4 necessary to produce 27.7g of H2 (g) according to the following reaction? 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)

mols H2 you want. grams/molar mass

Convert mols H2 to mols H2SO4 using the coefficients in the balanced equation.
Then M H2SO4 = mols H2SO4/L H2SO4. You know mols and M, solve for L and convert to mL if desired.

To find the minimum amount of 5.9M H2SO4 necessary to produce 27.7g of H2 according to the given reaction, we need to set up a stoichiometry calculation.

Let's start by finding the molar mass of H2. Hydrogen gas (H2) has a molar mass of approximately 2 g/mol.

Next, let's convert the given mass of H2 (27.7g) to moles. We can use the molar mass of H2 to do this calculation.

27.7g H2 × (1 mol H2 / 2 g H2) = 13.85 mol H2

According to the balanced equation, the stoichiometric ratio between H2 and H2SO4 is 3:3. It means that for every 3 moles of H2 produced, we need 3 moles of H2SO4.

So, we can calculate the moles of H2SO4 using the ratio:

13.85 mol H2 × (3 mol H2SO4 / 3 mol H2) = 13.85 mol H2SO4

Now, we know that 13.85 moles of H2SO4 are required to produce 27.7g of H2.

To find the minimum amount of 5.9M H2SO4, we need to calculate the volume (in liters) of 5.9M H2SO4 that contains 13.85 moles.

The definition of molarity (M) is moles of solute divided by the volume of the solution in liters:

M = mol solute / L solution

Rearranging the equation and plugging in the given values:

(5.9 mol H2SO4 / 1 L H2SO4) = (13.85 mol H2SO4 / V L H2SO4)

Solving for V (volume):

V = (13.85 mol H2SO4) / (5.9 mol/L)

V ≈ 2.35 L

Therefore, the minimum amount of 5.9M H2SO4 necessary is approximately 2.35 liters.