How many moles of Al(OH) are required to completely react with 2.75mol of H2SO4?
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.
To determine the number of moles of Al(OH)3 required to completely react with 2.75 mol of H2SO4, we need to examine the balanced chemical equation for the reaction between Al(OH)3 and H2SO4:
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.