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Answer 1

Question:

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

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

Answer:

1) KOH

2) Al(NO₃)₃

3) 0.01333 moles or 1.04004 g of Al(OH)₃

4) 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) The percentage yield of Al(OH)₃ is 48.075%

Explanation:

1) The reaction is given as follows;

Al(NO₃)₃ + 3KOH  → Al(OH)₃ + 3KNO₃

One mole of Al(NO₃)₃  reacts with three moles of 3KOH  to produce one mole of Al(OH)₃ and three moles of 3KNO₃

Hence when 0.4 moles of Al(NO₃)₃ reacts with 0.4 moles of KOH, the limiting reagent is found as follows;

1/3 moles Al(NO₃)₃  combines with 1 moles  of KOH

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;

$Percentage \ yield = \frac{Actual \, yield}{Theoretical \, yield} \times 100 = \frac{0.5}{1.04004} \times 100 = 48.075 \%$

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