If 8.42 × 10^{− 2}moles of aluminum react with 1.77 × 10^{− 2}moles of chlorine gas, how many moles of AlCl₃ can form?

Name: If 8.42 × 10 − 2 moles of aluminum react with 1.77 × 10 − 2 moles of chlorine gas, how many moles of AlCl 3 can form?
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Question

Quizplus · Accepted Answer

The balanced reaction is 2Al + 3Cl2 → 2AlCl3. The limiting reactant is Cl2, and using stoichiometry, 1.77 × 10^−2 moles of Cl2 will produce 1.18 × 10^−2 moles of AlCl3.

If 842 × 10 − 2moles of Aluminum React with 1

If 842 × 10^{− 2}moles of Aluminum React with 1