Answer:
71.372 g or 0.7 moles
Explanation:
We are given;
- Moles of Aluminium is 1.40 mol
- Moles of Oxygen 1.35 mol
We are required to determine the theoretical yield of Aluminium oxide
The equation for the reaction between Aluminium and Oxygen is given by;
4Al(s) + 3O₂(g) → 2Al₂O₃(s)
From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.
Therefore;
1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen
1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium
Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.
4 moles of aluminium reacts to generate 2 moles aluminium oxide.
Therefore;
Mole ratio Al : Al₂O₃ is 4 : 2
Thus;
Moles of Al₂O₃ = Moles of Al × 0.5
= 1.4 moles × 0.5
= 0.7 moles
But; 1 mole of Al₂O₃ = 101.96 g/mol
Thus;
Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol
= 71.372 g
I will say this is True….?
The equation for this question could be → .
so for 6.75 moles of *3 moles of O_{2}/2 moles of
= 10.125
Answer:
Option D. Al is above H on the activity series.
Explanation:
The equation for the reaction is given below:
2Al + 6HBr —> 2AlBr₃ + 3H₂
The activity series gives us a background understanding of the reactivity of elements i.e how elements displace other elements when present in solution.
From the activity series of metals, we understood that metal higher in the series will displace those lower in the series.
Considering the equation given above, Al is higher than H in the activity series. Thus, the reaction will proceed as illustrated by the equation.
Therefore, we can conclude that the reaction will only occur if Al is higher than H in the activity series.
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