Asked by Veronica
2Al(s)+3Cl2 = 2AlCl3
Limiting Reactant is Cl with 0.329mol.
What is the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine?
Limiting Reactant is Cl with 0.329mol.
What is the maximum mass of aluminum chloride that can be formed when reacting 30.0g of aluminum with 35.0g of chlorine?
Answers
Answered by
Devron
35g of Al*(1mole of Cl/35.45g)= mole of Cl
3 moles of Cl= 2 moles of AlCl3
moles of Cl *(2 moles of AlCl3/3 moles of Cl)= moles of AlCl3
moles of ALCl3 *(133.34 g of AlCl3/mol)= mass of AlCl3
3 moles of Cl= 2 moles of AlCl3
moles of Cl *(2 moles of AlCl3/3 moles of Cl)= moles of AlCl3
moles of ALCl3 *(133.34 g of AlCl3/mol)= mass of AlCl3
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.