Chemistry, asked by rabbianaeem121, 9 months ago

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

Answered by Anonymous
13

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

Answered by VaibhavSR
0

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.

Calculate the number of grams of k2so4 and water produced when 14g of KOH are reacted with excess of H2SO4

https://brainly.in/question/19681184

Calculate the number of grams of Al in 371 gr of Al2O3

https://brainly.in/question/47683655

#SPJ2

Similar questions