Question

3Al + 3 NH4ClO4 ---> Al2O3 + AlCl3 + 3 NO + 6H20

Answer 1

Answer:

9.63 L of NO

Explanation:

We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:

Mass of NH₄ClO₄ = 50 g

Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)

= 14 + 4 + 35.5 + 64

= 117.5 g/mol

Mole of NH₄ClO₄ =?

Mole = mass /molar mass

Mole of NH₄ClO₄ = 50/117.5

Mole of NH₄ClO₄ = 0.43 mole

Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:

3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O

From the balanced equation above,

3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.

Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.

Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:

1 mole of NO = 22.4 L

Therefore,

0.43 mole of NO = 0.43 × 22.4

0.43 mole of NO = 9.63 L

Thus, 9.63 L of NO were obtained from the reaction.