For the equation below, if 58.6 grams of Al were to react with 540 grams of H2SO4, how many grams of H2 would be produced?

2Al+3H2SO4--->1Al2(SO4)3+3H2

To determine the number of grams of H2 produced when 58.6 grams of Al reacts with 540 grams of H2SO4, you need to follow these steps:

Step 1: Convert the given masses of Al and H2SO4 to moles using their molar masses.
- Molar mass of Al: 26.98 g/mol
- Molar mass of H2SO4: 98.09 g/mol

Moles of Al = (58.6 g) / (26.98 g/mol)
Moles of H2SO4 = (540 g) / (98.09 g/mol)

Step 2: Determine the limiting reactant.
To determine the limiting reactant, you need to compare the mole ratios of Al to H2SO4 in the balanced chemical equation.
From the equation: 2 moles of Al react with 3 moles of H2SO4.

Moles of Al needed = 2 * (moles of H2SO4)

If the moles of Al available are greater than this amount, Al is in excess. If the moles of Al available are less than this amount, H2SO4 is in excess.
Compare the calculated moles of Al to the moles needed.

Step 3: Calculate the moles of H2 produced based on the limiting reactant.
In this case, since we have determined that H2SO4 is the limiting reactant, we will use its moles to calculate the moles of H2 produced.
From the balanced chemical equation, the mole ratio of H2 to H2SO4 is 3 moles of H2SO4 to 3 moles of H2.

Moles of H2 = 3 * (moles of H2SO4)

Step 4: Convert the moles of H2 to grams using its molar mass.
- Molar mass of H2: 2.02 g/mol

Grams of H2 = (moles of H2) * (2.02 g/mol)

By following these steps, you should be able to calculate the grams of H2 produced when 58.6 grams of Al reacts with 540 grams of H2SO4.

To find the number of grams of H2 that would be produced in the reaction, you need to use stoichiometry, which is the calculation of the quantities of reactants and products in chemical reactions.

First, write down the balanced equation for the reaction:
2Al + 3H2SO4 → 1Al2(SO4)3 + 3H2

From the balanced equation, you can see that 2 moles of Al react with 3 moles of H2SO4 to produce 3 moles of H2. This can be written as the following stoichiometric ratio:
2 moles Al : 3 moles H2SO4 : 3 moles H2

To determine the number of moles of H2, you need to convert grams of Al and grams of H2SO4 to moles using their respective molar masses.

The molar mass of Al is 26.98 grams/mol, so 58.6 grams of Al is equal to:
58.6 g Al × (1 mol Al / 26.98 g Al) = 2.17 moles of Al

The molar mass of H2SO4 is 98.09 grams/mol, so 540 grams of H2SO4 is equal to:
540 g H2SO4 × (1 mol H2SO4 / 98.09 g H2SO4) = 5.51 moles of H2SO4

Now, you can use the stoichiometric ratio to find the number of moles of H2. Since the ratio is 2 moles of Al to 3 moles of H2, you can write the following proportion:
2.17 moles Al : 3 moles H2SO4 = 3 moles H2 : x moles H2

Solving for x, you can cross-multiply and then divide:
(2.17 moles Al)(3 moles H2) = (3 moles H2SO4)(x moles H2)
6.51 = 3x
x = 6.51 / 3
x ≈ 2.17 moles of H2

To convert moles of H2 to grams, you can use the molar mass of H2, which is 2.02 grams/mol:
2.17 moles H2 × (2.02 g H2 / 1 mol H2) ≈ 4.38 grams of H2

Therefore, approximately 4.38 grams of H2 would be produced in the reaction.