QuestionJune 7, 2025

Consider the reaction below between Chlorine gas and Aluminum: 3Cl_(2)(g)+2Al(s)- 2AlCl_(3)(s) 15 grams of Cl_(2) is reacted with 10 grams of Al. How many grams of AlCl_(3) will be produced?

Consider the reaction below between Chlorine gas and Aluminum: 3Cl_(2)(g)+2Al(s)- 2AlCl_(3)(s) 15 grams of Cl_(2) is reacted with 10 grams of Al. How many grams of AlCl_(3) will be produced?
Consider the reaction below between Chlorine gas and Aluminum:
3Cl_(2)(g)+2Al(s)- 2AlCl_(3)(s)
15 grams of Cl_(2) is reacted with 10 grams of Al.
How many grams of AlCl_(3) will be produced?

Solution
4.7(389 votes)

Answer

18.8 grams of AlCl_3 will be produced. Explanation 1. Calculate moles of Cl_2 Molar mass of Cl_2 is 70.9 g/mol. Moles = \frac{15 \text{ g}}{70.9 \text{ g/mol}} = 0.2115 mol. 2. Calculate moles of Al Molar mass of Al is 26.98 g/mol. Moles = \frac{10 \text{ g}}{26.98 \text{ g/mol}} = 0.3707 mol. 3. Determine limiting reactant From the balanced equation, 3 moles of Cl_2 react with 2 moles of Al. Ratio for Cl_2 is \frac{0.2115}{3} = 0.0705, and for Al is \frac{0.3707}{2} = 0.18535. Cl_2 is the limiting reactant. 4. Calculate moles of AlCl_3 produced From the balanced equation, 3 moles of Cl_2 produce 2 moles of AlCl_3. Moles of AlCl_3 = 0.2115 \times \frac{2}{3} = 0.141 mol. 5. Calculate grams of AlCl_3 Molar mass of AlCl_3 is 133.34 g/mol. Grams = 0.141 \text{ mol} \times 133.34 \text{ g/mol} = 18.8 g.

Explanation

1. Calculate moles of $Cl_2$<br /> Molar mass of $Cl_2$ is 70.9 g/mol. Moles = $\frac{15 \text{ g}}{70.9 \text{ g/mol}} = 0.2115$ mol.<br />2. Calculate moles of Al<br /> Molar mass of Al is 26.98 g/mol. Moles = $\frac{10 \text{ g}}{26.98 \text{ g/mol}} = 0.3707$ mol.<br />3. Determine limiting reactant<br /> From the balanced equation, 3 moles of $Cl_2$ react with 2 moles of Al. Ratio for $Cl_2$ is $\frac{0.2115}{3} = 0.0705$, and for Al is $\frac{0.3707}{2} = 0.18535$. $Cl_2$ is the limiting reactant.<br />4. Calculate moles of $AlCl_3$ produced<br /> From the balanced equation, 3 moles of $Cl_2$ produce 2 moles of $AlCl_3$. Moles of $AlCl_3$ = $0.2115 \times \frac{2}{3} = 0.141$ mol.<br />5. Calculate grams of $AlCl_3$<br /> Molar mass of $AlCl_3$ is 133.34 g/mol. Grams = $0.141 \text{ mol} \times 133.34 \text{ g/mol} = 18.8$ g.
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