calculate the number of grams of AL2S3 which can be prepared by the reaction of 20g of (Al) and 30g of (sulphur).How much the non_limiting reactant is in excess?
Answers
Answer:
20g of Al = 20/26.98 mol = 0.74 mol
30g of S = 30/32.07 mol = 0.94 mol
so, only min(.74/2,.94/3) = 0.31333 mol of Al2S3 can be formed:
.31333 mol Al2S3 = .61667*26.98g Al + .94*32.07 g S = 16.64g Al + 30g S
That leaves 3.36g Al unreacted
Answer:
That leaves 3.36g Al unreacted.
Explanation:
Concept
The restricting reagent in a chemical response is the reactant with the intention to be fed on completely. Once there may be no extra of that reactant, the response can't proceed. Therefor it limits the response from continuing. The extra reagent is the reactant that would preserve reacting if the alternative had now no longer been fed on.
Given
AL₂S₃ which may be organized with the aid of using the response of 20g Al and 30g of sulphur
To find
calculate the variety of grams of AL₂S₃
Solution
20g of Al = 20/26.ninety eight mol = 0.seventy four mol
30g of S = 30/32.07 mol = 0.ninety four mol
so, best min(.seventy four/2,.ninety four/3) = 0.31333 mol of Al₂S₃ may be formed:
.31333 mol Al₂S₃ = .61667 X 26.98g Al + .ninety four X 32.07 g S = 16.64g Al + 30g S
That leaves 3.36g Al unreacted.
The very last answer
That leaves 3.36g Al unreacted.
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