Question

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?

Answers

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

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