Question

If 1.20 g of aluminum hydroxide reacts with 3.00 g of sulfuric acid, what is the mass of water produced?

Al(OH)3(s)+H2SO4(l)→Al2(SO4)3(aq)+H2O(l)

Answer 1

To calculate the mass of water produced when 1.20 g of aluminum hydroxide reacts with 3.00 g of sulfuric acid, we need to determine the limiting reactant and use the mole ratio from the balanced chemical equation.

Step-by-step explanation:

To calculate the mass of water produced when 1.20 g of aluminum hydroxide reacts with 3.00 g of sulfuric acid, we first need to determine the limiting reactant. The balanced chemical equation tells us that the mole ratio of Al(OH)3 to H2O is 1:1. We can convert the masses of Al(OH)3 and H2SO4 to moles using their molar masses, and then compare the moles to determine the limiting reactant. Once we know the limiting reactant, we can find the moles of H2O produced and convert it back to grams using the molar mass of water.

First, let's find the moles of Al(OH)3 and H2SO4:

Moles of Al(OH)3 = 1.20 g / (molar mass of Al(OH)3)

Moles of H2SO4 = 3.00 g / (molar mass of H2SO4)

Next, let's compare the moles to determine the limiting reactant. The reactant with the smaller number of moles is the limiting reactant.

Once we know the limiting reactant, we can find the moles of H2O produced by using the mole ratio from the balanced chemical equation. Finally, we can convert the moles of H2O to grams:

Mass of H2O produced = Moles of H2O * (molar mass of H2O)

Answer 2

 it would be 342 acid