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100.0mL of 0.40M Al(NO3)3 is reacted with 100.0 mL of 0.40M Al(OH)3 according to

the following reaction: Al(NO3)3 + 3KOH → Al(OH)3 + 3KNO3

1) Identify the limiting reactant.

2) Identify the excess reactant.

3) Determine the theoretical yield of Al(OH)3

4) How much excess reactant is left over after the reaction is complete?

5) If 0.50g of Al(OH)3 determine the percent yield of Al(OH)3

Hence 0.4/3 moles of Al(NO₃)₃ will combine with 0.4 moles of KOH from which there will be an excess of (0.4 - 0.4/3) moles of Al(NO₃)₃ left, making KOH the limiting reactant

2) The excess reactant as shown above is the Al(NO₃)₃

3) From the reaction above, 0.4 moles of KOH combined with excess Al(NO₃)₃ will produce 0.4/3 moles of Al(OH)₃ however, since the reagent are each 100 mL, and 0.40 M indicates 0.4 moles per 1000 mL the number of moles of Al(OH)₃ produced is given as follows;

0.4/3 = 2/15 moles per 1000 mL ≡ 2/15 ÷ 10 = 2/150 moles per 1000 mL ÷ 10 or 100 mL

The number of moles of Al(OH)₃ produced = 2/150 = 0.01333 moles

Molar mass of Al(OH)₃ = 78.003 g/mol

Mass of Al(OH)₃ produced = Number of moles produced × Molar mass

Mass of Al(OH)₃ produced = 0.01333 × 78.003 = 1.04004 g

4) The amount of the excess reactant, Al(NO₃)₃ left after the reaction is found as follows;

0.4 - 0.4/3 = 2/5 - 2/15 = 4/15 moles of Al(NO₃)₃ per 1000 mL or 4/(15×10) =2/75 moles per 100 mL

Molar mass of Al(NO₃)₃ 212.996238 g/mol

Mass left = Number of moles × Molar mass = 5.6799 grams

The amount of the excess reactant, Al(NO₃)₃ left after the reaction = 2/75 moles or 5.6799 grams of Al(NO₃)₃

5) If 0.50 g of Al(OH)₃ is produced, we have

Actual yield = 0.01333 moles of Al(OH)₃

Molar mass of Al(OH)₃ = 78.003 g/mol

Mass of theoretical yield = Molar Mass × Theoretical yield = 0.01333 × 78.003 = 1.04004 g of Al(OH)₃

Therefore;

[tex]Percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{0.5}{1.04004} \times 100 = 48.075 \%[/tex]

The percentage yield of Al(OH)₃ = 48.075%.