Question

11. Write the balanced equation for the reaction of HC2H3O2, with Al(OH)3 to form H2O and Al(C2H302)3:

a. Using the equation from problem #11, determine the mass of aluminum acetate that can be
made if you do this reaction with 125 grams of acetic acid (HC,H,O2).
b. Using the equation from problem #11, determine the mass of aluminum acetate that can be
made if you do this reaction with 275 grams of aluminum hydroxide?

Answer 1

74.36 g of aluminium acetate.

730.27g of aluminium acetate.

- to the nearest hundredth.

Explanation:

Acetic acid is usually written as CH3COOH.

a. 6CH3COOH + Al(OH)3 ---> Al(CH3COO)3 + 9H2O

So 6 moles of acetic acid produce 1 mole of aluminium acetate.

Using the molecular masses

6*( 1.008*4 + 12.011*2 + 16 *2) g acetic acid gives (26.98+3(36.032+ 2*12.011)

348.228 g acetic acid gives 207.142 g Al acetate.

So 125 g gives (207.142 / 348.228) * 125

= 74.36 g of aluminium acetate.

b.

(26.98 + 3*16 + 3 * 1.008) g of Al(OH)3 gives 207.142 g Al acetate

78.004 g gives 207.142 g Al acetate

275 g gives (207.142 / 78.004) * 275

= 730.27g Al acetate.