Answer:
1.795 mole of H2.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2Al + 6HCl —> 2AlCl3 + 3H2
Step 2:
Determination of the limiting reactant.
From the balanced equation above,
2 moles of Al reacted with 6 moles
Therefore, 2.57 moles of Al will react with = (2.57 x 6)/2 = 7.71 moles of HCl.
From the calculation made above, it will require a higher amount of HCl than what was given to react completely with 2.57 moles of Al. Therefore, HCl is the limiting reactant and Al is the excess reactant.
Step 3:
Determination of the number of mole H2 produced from the reaction.
Here, we shall be using the limiting reactant because it will produce the maximum yield of the reaction since all of it were consumed by the reaction.
The limiting reactant is HCl and the amount of H2 produce can be obtained as follow:
From the balanced equation above,
6 moles of HCl reacted to produce 3 moles of H2.
Therefore, 3.59 moles of HCl will produce = (3.59 x 3)/6 = 1.795 mole of H2.
From the calculations made above, 1.795 mole of H2 is produced from the reaction.
