For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants.
2Al(s) +3 Cl2(g) ---> 2 AlCl3(s)
A. 2.5gAl , 2.5g Cl2
B. 7.7 g Al , 25.2 g Cl2
C. 0.240 g Al , 1.10 g Cl2
duplicate post. See original response below.
To calculate the theoretical yield of the product (AlCl3), we need to determine the limiting reactant in each case and then use stoichiometry to find the amount of product formed.
First, let's calculate the number of moles for each reactant:
A. 2.5g Al
Molar mass of Al = 26.98 g/mol
Number of moles of Al = mass / molar mass = 2.5 g / 26.98 g/mol = 0.093 mol
2.5g Cl2
Molar mass of Cl2 = 35.45 g/mol
Number of moles of Cl2 = mass / molar mass = 2.5 g / 35.45 g/mol = 0.071 mol
B. 7.7 g Al
Number of moles of Al = 7.7 g / 26.98 g/mol = 0.285 mol
25.2g Cl2
Number of moles of Cl2 = 25.2 g / 35.45 g/mol = 0.711 mol
C. 0.240 g Al
Number of moles of Al = 0.240 g / 26.98 g/mol = 0.009 mol
1.10 g Cl2
Number of moles of Cl2 = 1.10 g / 35.45 g/mol = 0.031 mol
Next, we need to determine the limiting reactant. The reactant that produces the least amount of product is the limiting reactant.
A. Comparing the stoichiometry of the balanced equation, we can see that the molar ratio of Al to Cl2 is 2:3. Since the mole ratio is not balanced, we need to find the limiting reactant by comparing their stoichiometric ratios.
Mole ratio Al : Cl2 = 0.093 mol : 0.071 mol = 1.31 : 1
The ratio is closer to 1:1, indicating that Al is the limiting reactant. Therefore, the amount of AlCl3 produced will be determined by the moles of Al.
Molar mass of AlCl3 = 133.34 g/mol
Theoretical yield of AlCl3 = number of moles of Al x molar mass of AlCl3 = 0.093 mol x 133.34 g/mol = 12.39 g
B. Mole ratio Al : Cl2 = 0.285 mol : 0.711 mol = 0.4 : 1
The ratio is closer to 0.4:1, indicating that Al is the limiting reactant. Therefore, the amount of AlCl3 produced will be determined by the moles of Al.
Theoretical yield of AlCl3 = 0.285 mol x 133.34 g/mol = 38.02 g
C. Mole ratio Al : Cl2 = 0.009 mol : 0.031 mol = 0.29 : 1
The ratio is closer to 0.29:1, indicating that Al is the limiting reactant. Therefore, the amount of AlCl3 produced will be determined by the moles of Al.
Theoretical yield of AlCl3 = 0.009 mol x 133.34 g/mol = 1.20 g
To sum up the theoretical yields for each case:
A. Theoretical yield = 12.39 g
B. Theoretical yield = 38.02 g
C. Theoretical yield = 1.20 g
To determine the theoretical yield of a product in a chemical reaction, you need to use stoichiometry, which is the study of the quantitative relationships between reactants and products in a chemical reaction.
Let's start by calculating the moles of each reactant based on their given masses and the molar mass of each substance. The molar mass can be found on the periodic table.
A. For 2.5 g Al:
Molar mass of Al = 26.98 g/mol
Moles of Al = (mass of Al / molar mass of Al) = (2.5 g / 26.98 g/mol)
For 2.5 g Cl2:
Molar mass of Cl2 = 70.91 g/mol
Moles of Cl2 = (mass of Cl2 / molar mass of Cl2) = (2.5 g / 70.91 g/mol)
B. Repeat the same steps as above using the given masses of Al and Cl2.
C. Repeat the same steps as above using the given masses of Al and Cl2.
Next, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
To determine the limiting reactant, compare the mole ratios of the reactants in the balanced chemical equation. In this case, the balanced equation is 2Al(s) + 3Cl2(g) -> 2AlCl3(s).
Calculate the mole ratio of Al and Cl2 using the coefficients from the balanced equation:
Mole ratio Al to Cl2 = 2 / 3
Compare the mole ratio to the actual mole ratio of Al to Cl2 calculated above for each case.
The reactant with the smaller mole ratio is the limiting reactant.
Now, we can calculate the theoretical yield of the product using the stoichiometry of the balanced equation.
To do this, determine the mole ratio between the limiting reactant and the product (AlCl3) using the balanced equation:
Mole ratio AlCl3 to Al = 2 / 2
Multiplying the moles of the limiting reactant (determined earlier) by the mole ratio of the product to the limiting reactant will give you the moles of the product formed.
Finally, convert the moles of the product to grams using the molar mass of the product.
Repeat these steps for each given initial amount of reactants to calculate the theoretical yield of the product for each case.