Answer :
Substances are consumed according to the coefficients in front of them.
0.32 mol of Al is consumed, so that means we use 0.08 mol per coefficient of one in front of the substances. Thus, we use 0.3 × 0.08 = 0.24 mol of O₂ gase. We do not use all of the O₂ gase, but in contrast, we use all of the Al. Hence, Al is the limiting reactant here. I personally divide the number of moles by the coefficient of the reactant and whichever of them is smaller, it's the limiting reactant.
Limiting reactant may change depending on how many moles we take from both reactants. Here, we have to determine which one is the limiting one. For simplification, we can ignore 10⁻³s.
For O₂: 6.38 ÷ 3 ≅ 2.12
For Al: 9.15 ÷ 4 ≅ 2.28
O₂ is the limiting reactant.
Reaction calculations are done according to the limiting reactants' being consumed. If 9.15 × 10⁻³ mole of O₂ is consumed and, of course its coefficient is 3, the amount of Al₂O₃, whose coefficient is 2, will be (2 ÷ 3) × 9.15 × 10⁻³ = 6.1 × 10⁻³ mol
We are given the masses of the substances. We have to find, firstly, their number of moles. Molar mass of Al is 27 gr. while that of O₂ is 18.
3.17 ÷ 27 will give us the number of moles of Al, which equals ≅ 0.12
2.55 ÷ 18 will give us the number of moles of O₂, which equals ≅ 0.14
We will divide the number of moles by the coefficients for both substances and the smaller one will be the limiting reactant.
For Al, ≈0.12 ÷ 4 ≅ 0.03
For O₂, ≈0.14 ÷ 3 ≅ 0.05
The smaller one is that of Al, so Al is the limiting reactant.
If there is anything that you have a difficulty understanding or you think anything is wrong, please let me know, everyone makes mistakes and I am sorry for my bad English ^^
The correct answer are as follows
- Al is the limiting reactant
- 4.25 moles of Al
- O2 is limiting reactant
What is Limiting reactant?
- Those reactants that get completely utilized in a reaction first and thus limit the amount of product that will be produced are known as limiting reactants.
How to calculate the limiting reactants?
Limiting reactant can be calculated by the following ways.
- Calculate the number of moles of each reactant present.
- Compare this ratio to the mole ratio of the reactants in the balanced chemical equation.
How to Solve these question?
- The stoichiometry of the reaction is such that 4 moles of Al are required for every 3 moles of diatomic oxygen.
- If the ratio of Al to diatomic oxygen is greater than 4/3 then the oxygen is the limiting reagent. If the ratio of Al to diatomic oxygen is less than 4/3, then Al is the limiting reagent.
1. The number of mole of Al given is 0.32
The number of mole of O2 given is 0.26
Now for Al,
(0.32mol Al) * (2mol Al2O3 /4mol Al) = 0.160mol Al2O3.
Now for O2
(0.26mol O2) * (2mol Al2O3 /3mol O2) = 0.173mol Al2O3
As we can see that there is not enough Al to consume O2
Hence Al is the Limiting reactant.
2. We need to find the number of moles of Al2O3
So here for O2
(6.38*10-3mol O2) * (2mol Al2O3 /3mol O2)
= 4.25*10-3mol Al2O3.
For Al
(9.15*10-3mol Al) * (2mol Al2O3 /4mol Al)
= 4.58*10-3mol Al2O3.
3. 3.17g of Al is given
2.55g of O2 is given
For Al
(3.17g Al) *(1mol Al /26.98g Al) * (2mol Al2O3 /4mol Al) = 0.0587mol Al2O3.
For O2
(2.55g O2) * (1mol O2 /32.0g O2) * (2mol Al2O3 / 3mol O2) = 0.0531mol Al2O3.
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