aluminum reacts with sulfuric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be formed if 250 g H2SO4 completly reacted with aluminum

To determine the number of grams of aluminum sulfate formed, we need to use the balanced chemical equation of the reaction.

The balanced equation for the reaction between aluminum and sulfuric acid is as follows:

2Al + 3H2SO4 → Al2(SO4)3 + 3H2

From the equation, we can see that for every 3 moles of H2SO4, we will form 1 mole of Al2(SO4)3.

Step 1: Calculate the moles of sulfuric acid (H2SO4) given 250g.

To calculate the number of moles, we need to divide the given mass by the molar mass of H2SO4. The molar mass of H2SO4 is:

Molar mass of H = 1 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16 g/mol

Molar mass of H2SO4 = (2 x 1) + 32.07 + (4 x 16) = 98.09 g/mol

Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4
Moles of H2SO4 = 250g / 98.09 g/mol ≈ 2.55 mol (rounded to two decimal places)

Step 2: Calculate the moles of Al2(SO4)3 formed.

From the balanced equation, we know that for every 3 moles of H2SO4, we will form 1 mole of Al2(SO4)3. Therefore, the moles of Al2(SO4)3 formed will be equal to one-third of the moles of H2SO4.

Moles of Al2(SO4)3 = 1/3 * Moles of H2SO4
Moles of Al2(SO4)3 = 1/3 * 2.55 mol ≈ 0.85 mol (rounded to two decimal places)

Step 3: Calculate the mass of Al2(SO4)3 formed.

To calculate the mass, we need to multiply the moles of Al2(SO4)3 formed by its molar mass. The molar mass of Al2(SO4)3 is:

Molar mass of Al2 = (2 x 26.98) = 53.96 g/mol
Molar mass of S = 32.07 g/mol
Molar mass of O = 16 g/mol

Molar mass of Al2(SO4)3 = (2 x 53.96) + (3 x 32.07) + (12 x 16) = 342.14 g/mol

Mass of Al2(SO4)3 = Moles of Al2(SO4)3 * Molar mass of Al2(SO4)3
Mass of Al2(SO4)3 = 0.85 mol * 342.14 g/mol ≈ 291.82 g (rounded to two decimal places)

Therefore, approximately 291.82 grams of aluminum sulfate would be formed if 250 grams of H2SO4 completely reacted with aluminum.

To determine the amount of aluminum sulfate formed when 250 g of sulfuric acid (H2SO4) reacts with aluminum, we need to first balance the chemical equation:

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

From the balanced equation, we can see that for every 2 moles of aluminum, we obtain 1 mole of aluminum sulfate. Therefore, we need to convert grams of H2SO4 to moles of H2SO4, and then use the mole ratio to find the moles of aluminum sulfate.

1. Convert grams of H2SO4 to moles:
To convert grams to moles, we need the molar mass of H2SO4. The molar mass of H2SO4 is calculated by summing the atomic masses of the elements in the compound.

H2SO4: (2 * 1.008 g/mol) + (1 * 32.07 g/mol) + (4 * 16.00 g/mol) = 98.09 g/mol

Now we can calculate the number of moles of H2SO4:
moles of H2SO4 = (mass of H2SO4) / (molar mass of H2SO4)
moles of H2SO4 = 250 g / 98.09 g/mol ≈ 2.55 mol H2SO4

2. Use the mole ratio to find the moles of aluminum sulfate:
From the balanced equation, the mole ratio between H2SO4 and Al2(SO4)3 is 3:1. Therefore, we multiply the moles of H2SO4 by the ratio.

moles of Al2(SO4)3 = (moles of H2SO4) * (1/3)
moles of Al2(SO4)3 = 2.55 mol * (1/3) ≈ 0.85 mol Al2(SO4)3

3. Convert moles of aluminum sulfate to grams:
To convert moles to grams, we need to use the molar mass of Al2(SO4)3.

Al2(SO4)3: (2 * 26.98 g/mol) + (3 * (32.07 g/mol + 4 * 16.00 g/mol)) = 342.14 g/mol

grams of Al2(SO4)3 = (moles of Al2(SO4)3) * (molar mass of Al2(SO4)3)
grams of Al2(SO4)3 = 0.85 mol * 342.14 g/mol ≈ 291.32 g Al2(SO4)3

Therefore, approximately 291.32 grams of aluminum sulfate would be formed when 250 grams of H2SO4 completely reacts with aluminum.

2Al + 3H2SO4 ==> Al2(SO4)2 + 3H2O

mols H2SO4 = grams/molar mass
Using the coefficients in the balanced equation, convert mols H2SO4 5omols Al2(SO4)3.
Now convert mols Al2(SO4)3 to grams. g = mols x molar mass.