The chemical equation for the reaction of acetic acid with aluminum hydroxide to form water and aluminum acetate is presented below:HC2H3O2 + Al(OH)3 --> Al(C2H3O2)3 + H2OHow many grams of HC2H3O2 are needed to react completely with 403 grams of Al(OH)3?Write your final answer in two decimal places.Use the following mass numbers:C - 12 g/molH - 1 g/molO - 16 g/molAl - 27 g/mol

First, we need to determine the molar mass of each compound involved in the reaction.

For acetic acid (HC2H3O2), the molar mass is calculated as follows: H: 1 g/mol * 4 = 4 g/mol C: 12 g/mol * 2 = 24 g/mol O: 16 g/mol * 2 = 32 g/mol Total molar mass of HC2H3O2 = 4 + 24 + 32 = 60 g/mol

For aluminum hydroxide (Al(OH)3), the molar mass is calculated as follows: Al: 27 g/mol * 1 = 27 g/mol O: 16 g/mol * 3 = 48 g/mol H: 1 g/mol * 3 = 3 g/mol Total molar mass of Al(OH)3 = 27 + 48 + 3 = 78 g/mol

From the balanced chemical equation, we can see that the molar ratio of HC2H3O2 to Al(OH)3 is 1:1. This means that one mole of acetic acid reacts with one mole of aluminum hydroxide.

Next, we need to convert the mass of Al(OH)3 given (403 g) to moles using its molar mass: 403 g Al(OH)3 * (1 mol Al(OH)3 / 78 g Al(OH)3) = 5.17 mol Al(OH)3

Since the molar ratio of HC2H3O2 to Al(OH)3 is 1:1, we need the same number of moles of acetic acid to react completely with the aluminum hydroxide. Therefore, we need 5.17 moles of HC2H3O2.

Finally, we convert this amount in moles back to grams using the molar mass of acetic acid: 5.17 mol HC2H3O2 * (60 g HC2H3O2 / 1 mol HC2H3O2) = 310.2 g HC2H3O2

So, 310.20 grams of HC2H3O2 are needed to react completely with 403 grams of Al(OH)3.