How many grams of aluminum sulfate Al2(SO4)3 , will be produced from the complete reaction of 270g of Al

3 Answers | Add Yours

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`2Al_((s))+3H_2SO_(4(aq)) rarr Al_2(SO_4)_(3(aq))+3H_(2(g))`

`Al_2(SO_4)_3` is formed when we add `H_2SO_4` acid to Al solid.

Mole ratio

`Al:Al_2(SO_4)_3 = 2:1`

Molar mass of Al = 27g/mol

Amount of Al reacted `= 270/27 = 10mol`

Amount of `Al_2(SO_4)_3` formed `= 1/2xx10 = 5mol`

Molar mass of `Al_2(SO_4)_3` = 342g/mol

Mass of `Al_2(SO_4)_3` obtained `= 342xx5 = 1710g`

So we can get a maximum of 1710g of `Al_2(SO_4)_3` .

Assumption

  • There is enough Sulphuric acid for the complete reaction
Sources:
Yojana_Thapa's profile pic

Yojana_Thapa | Student, Grade 10 | (Level 1) Valedictorian

Posted on

rams of aluminum sulfate , will be produced from the complete reaction of

xgAl2(SO4)3 / (54+ 96+192) = 270gAl / 2 x 27

- now we cross multiply

(54+96+192) = 342

342 x 270gAL= 92340

92340/ (2x27) = 1710

xgAl2(SO4)3 = 1710g

1710g grams of aluminum sulfate Al2(SO4)3 will be produced.

We’ve answered 318,967 questions. We can answer yours, too.

Ask a question