compute theoretical yield (g)
~2Al+3Cl2->2AlCl3
~2g Al, 2g Cl2
[ans:2.5g AlCl3]
i need the calculation work bcus i got diff from the ans which is 7.12 g.
moles Al = 2/26.98 = ??
moles Cl2 =2/70.9 = ??
moles Al x (2moles AlCl3/2 moles Al) = 0.0741 OR
moles Cl2 x (2 moles AlCl3/3 moles Cl2) = 0.0188
Therefore, the limiting reagent is Cl2, moles AlCl3 formed is 0.0188 and that x molar mass of approximately 133 = 2.5 g AlCl3 as theoretical yield.
ME TOO. i was doing this question and got 7.4!! im so confused
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ちょっと宣伝するのやめてもらえませんか？迷惑なので...
To calculate the theoretical yield of a chemical reaction, you need to determine which reactant is limiting and use stoichiometry to calculate the maximum amount of product that can be formed.
In the given reaction: 2Al + 3Cl2 -> 2AlCl3, the stoichiometric ratio between Al and AlCl3 is 2:2, which means for every 2 moles of Al, we will obtain 2 moles of AlCl3.
First, we need to determine which reactant is limiting, meaning which reactant will be completely consumed and restrict the formation of the product. This can be done by comparing the number of moles of each reactant to the stoichiometric ratio.
Given that the mass of Al is 2g and the molar mass of Al is approximately 27 g/mol, we can calculate the number of moles of Al as follows:
moles of Al = mass of Al / molar mass of Al
= 2g / 27 g/mol
≈ 0.074 mol
Similarly, the number of moles of Cl2 can be calculated:
moles of Cl2 = mass of Cl2 / molar mass of Cl2
= 2g / (2 * 35.5 g/mol)
≈ 0.056 mol
From the balanced equation, we can see that the stoichiometric ratio between Al and Cl2 is 2:3. Therefore, for every 2 moles of Al, we need 3 moles of Cl2.
To determine the limiting reactant, we compare the ratios of moles of Al to Cl2 with the stoichiometric ratio:
(0.074 mol Al) / 2 ≈ 0.037 and (0.056 mol Cl2) / 3 ≈ 0.019
The smaller value is 0.019, indicating that Cl2 is the limiting reactant.
Now, we need to calculate the theoretical yield of AlCl3 using the limiting reactant.
Using the mole ratio from the balanced equation, we know that for every 3 moles of Cl2, we will produce 2 moles of AlCl3.
moles of AlCl3 produced = moles of Cl2 * (2 moles AlCl3 / 3 moles Cl2)
= 0.056 mol * (2/3)
≈ 0.037 mol
To calculate the mass of AlCl3, we can use the molar mass of AlCl3, which is approximately 27 + (3 * 35.5) g/mol = 133.5 g/mol.
mass of AlCl3 = moles of AlCl3 * molar mass of AlCl3
= 0.037 mol * 133.5 g/mol
≈ 4.94 g
Hence, the theoretical yield of AlCl3 is approximately 4.94 g, which is different from the given answer of 2.5 g.
If you obtained a different answer (7.12 g) as mentioned, it is possible that there was an error in the calculation or the given mass of the reactants may have been incorrect. Please double-check your calculations and the given data to determine the cause of the discrepancy.