In what reaction? This one?
Al(OH)3+ 3H2SO4 >>Al2(SO4)3 + 3H2O
you need 1/3 the moles, or 2.75/3 moles of aluminum hydroxide.
Al(OH)3+ 3H2SO4 >>Al2(SO4)3 + 3H2O
you need 1/3 the moles, or 2.75/3 moles of aluminum hydroxide.
2 Al(OH)3 + 3 H2SO4 -> Al2(SO4)3 + 6 H2O
From the equation, we can see that the stoichiometric ratio between Al(OH)3 and H2SO4 is 2:3. This means that for every 2 moles of Al(OH)3, we need 3 moles of H2SO4 to react completely.
To find the moles of Al(OH)3 required to react with 2.75 mol of H2SO4, we can set up a proportion:
2 mol Al(OH)3 / 3 mol H2SO4 = x mol Al(OH)3 / 2.75 mol H2SO4
Cross-multiplying the proportion gives:
2 mol Al(OH)3 * 2.75 mol H2SO4 = 3 mol H2SO4 * x mol Al(OH)3
5.50 mol Al(OH)3 = 3x mol Al(OH)3
Now, divide both sides of the equation by 3, to solve for x:
x = 5.50 mol Al(OH)3 / 3
x ≈ 1.83 mol Al(OH)3
Therefore, approximately 1.83 moles of Al(OH)3 are required to completely react with 2.75 moles of H2SO4.