aluminum reacts with sulfric acid to produce aluminum sulfate and hydrogen gas. how many grams of aluminum sulfate would be form if 250 g H2SO4 completely reacted with aluminum?
To find out how many grams of aluminum sulfate would be formed when 250 g of H2SO4 completely reacts with aluminum, we need to calculate the stoichiometry of the reaction.
The balanced chemical equation for the reaction is:
2 Al + 3 H2SO4 → Al2(SO4)3 + 3 H2
From the equation, we can see that 3 moles of H2SO4 react with 1 mole of Al2(SO4)3.
To determine the number of moles of H2SO4 in 250 g, we need to use the molar mass of H2SO4, which is approximately 98 g/mol.
Number of moles of H2SO4 = Mass / Molar mass
Number of moles of H2SO4 = 250 g / 98 g/mol
Number of moles of H2SO4 ≈ 2.55 mol
From the stoichiometry, we know that 3 moles of H2SO4 react with 1 mole of Al2(SO4)3. Therefore, the number of moles of Al2(SO4)3 formed will be one-third of the moles of H2SO4.
Number of moles of Al2(SO4)3 = (1/3) * Number of moles of H2SO4
Number of moles of Al2(SO4)3 ≈ (1/3) * 2.55 mol
Number of moles of Al2(SO4)3 ≈ 0.85 mol
Now, to find the mass of aluminum sulfate formed, we need to use the molar mass of Al2(SO4)3, which is approximately 342 g/mol.
Mass of Al2(SO4)3 = Number of moles * Molar mass
Mass of Al2(SO4)3 ≈ 0.85 mol * 342 g/mol
Mass of Al2(SO4)3 ≈ 290.7 g
Therefore, approximately 290.7 grams of aluminum sulfate would be formed if 250 g of H2SO4 completely reacted with aluminum.
balance the equation FIRST
2Al+ 3H2SO4>>Al2(SO4)3 + 3H2
so, for each mole of acid, you get 1/3 mole of aluminum sulfate.
How many moles of sulfuric acid is 250grams?