For the given reaction, how many moles of water will be formed starting with 45.0 moles of aluminum hydroxide:

2AL(OH)3(s) + 3 H2SO4 > Al2(SO4)3 +
6H2O

Follow the steps in the example.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the number of moles of water formed, we need to use stoichiometry, which is a relationship between the quantities of substances involved in a chemical reaction.

First, let's calculate the moles of water formed by using the balanced equation:

2 Al(OH)3(s) + 3 H2SO4 -> Al2(SO4)3 + 6 H2O

From the equation, we can see that for every 2 moles of aluminum hydroxide (Al(OH)3), 6 moles of water (H2O) are formed.

Therefore, we can set up a ratio using the coefficients in the balanced equation:

2 moles of Al(OH)3 : 6 moles of H2O

Now, to determine the moles of water formed, we need to find the proportionality constant. We can do this by setting up a simple proportion:

(45.0 moles of Al(OH)3) / (2 moles of Al(OH)3) = (x moles of H2O) / (6 moles of H2O)

Cross-multiplying this proportion will give us:

(45.0 moles of Al(OH)3) × (6 moles of H2O) = (x moles of H2O) × (2 moles of Al(OH)3)

(x moles of H2O) = (45.0 moles of Al(OH)3) × (6 moles of H2O) / (2 moles of Al(OH)3)

Calculating this expression will give us the answer:

(x moles of H2O) = (45.0 moles of Al(OH)3) × (6 moles of H2O) / (2 moles of Al(OH)3)

(x moles of H2O) = (270 moles of H2O) / (2 moles of Al(OH)3)

(x moles of H2O) = 135 moles of H2O

Therefore, starting with 45.0 moles of aluminum hydroxide, 135 moles of water will be formed in the given reaction.