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.
