I am just confused with how to do these problems I cannot figure out how to set up the problem correctly. I have balanced out the equation as I know that this is the first step to solving these problems. Please can you explain how to do this my teacher taught us how to do it with a chart but I prefer Dimensional Analysis.
3A(g) + 2B(g) = ____ A3B2(g)
Considering the limiting reactant concept, how many moles of A3B2 are produced from the reaction of 5.0 mol A and 4.00 mol B?
2017-11-06 1:22 pm
回答 (2)
2017-11-06 1:49 pm
3A(g) + 2B(g) → A₃B₂(g)
Mole ratio A : B : A₃B₂ = 3 : 2 : 1
When 5.0 mol of A completely reacted :
(5.0 mol A) × (2 mol B/3 mol A) = 3.33 mol B < 4.00 mol B
Hence, B is in excess, and A is the limiting reactant which completely reacts.
(5.0 mol A) × (1 mol A₃B₂/ 3 mol A) = 1.67 mol A₃B₂ (produced)
Mole ratio A : B : A₃B₂ = 3 : 2 : 1
When 5.0 mol of A completely reacted :
(5.0 mol A) × (2 mol B/3 mol A) = 3.33 mol B < 4.00 mol B
Hence, B is in excess, and A is the limiting reactant which completely reacts.
(5.0 mol A) × (1 mol A₃B₂/ 3 mol A) = 1.67 mol A₃B₂ (produced)
2017-11-06 1:41 pm
5.0 mol of A would react completely with 5.0 x (2/3) = 3.33 mol of B, but there is more B present than that, so B is in excess and A is the limiting reactant.
(5.0 mol A) x (1 mol A3B2 / 3 mol A) = 1.7 mol A3B2
(5.0 mol A) x (1 mol A3B2 / 3 mol A) = 1.7 mol A3B2
收錄日期: 2021-05-01 21:56:45
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