Chemistry Help Please

Question

3Al (s) + 3NH₄ClO₄ -->Al₂O₃(s) + AlCl₃ (s) + 3NO (g) + 6H₂O (l)

1)Which reactant is the limiting reactant if 4.25 g of Al is mixed with 5.46 g of NH₄ClO₄

2)How many grams of Al₂O₃ can be produced from the conditions above ( theoretical yield)

3) what is the actual yield of Al₂O₃ in grams if the percent yield is 68.2%

Answer 1

Hello, E. P.,

Calculate the moles of both reactants by dividing their masses by their molar masses.

Moles

NH₄ClO₄ 1.34

Al 0.158

The balanced equation says we need equal molar amounts of both reactants. This will make the NH₄ClO₄ the limiting reagent.

To find moles Al₂O₃ theoretically produced, we note that the equation says we'll obtain 1 mole of Al₂O₃ for every 3 moles NH₄ClO₄ That's a molar ratio of (1 mole Al2O3)/(3 moles NH₄ClO4).

(1.34 moles H4ClO4)*(1 mole Al2O3)/(3 moles NH4ClO4) = 0.333 moles Al2O3, or 33.99 grams.  

At 68.2% yield, this is reduced to 23.18 grams Al2O3,

Bob