A) 2.5g Al, 2.5g Cl2 Express your answer using two significant figures. B) 7.5g

Question

A) 2.5g Al, 2.5g Cl2

Express your answer using two significant figures.

B) 7.5g Al, 25.2g Cl2

Express your answer using three significant figures.

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For the reaction shown, compute the theoretical yield of product (in grams) for each of the following initial amounts of reactants. 2Al(s)+3Cl2(g) right arrow 2AlCl3(s) A) 2.5g Al, 2.5g Cl2 Express your answer using two significant figures. B) 7.5g Al, 25.2g Cl2 Express your answer using three significant figures. open the picture in a new tap so you can see it clearly

Explanation / Answer

2 Al(s) + 3 Cl2(g) --> 2 AlCl3 (s)
molar mass of Al = 27, m.m. of Cl = 35.5, Therefore m.m of Cl2 = 71

So,
a) 2.5 g Al = (2.5/27) moles = 0.0926 moles
2.5g Cl2 = (2.5/71) = 0.0352 moles
3 moles of Cl2 react with 2 moles Al to produce 2 moles AlCl3
Therefore, 1 mole of Cl2 reacts with 2/3 moles of Al to produce 2/3 moles AlCl3
0.0352 moles Cl2 reacts with 0.0352 * 2/3 (= 0.0234) moles of Al to produce same amount (0.0234 moles) of AlCl3
THEREFORE some of the Al is left unreacted.
molar mass of AlCl3 = 133.5 g
yield of product in grams = molar mass * no of moles = 133.5 * 0.0234 = 3.133 g of AlCl3

b) Same as Above
7.5 g Al = 0.278 moles Al
25.2 g Cl2 = 0.354 moles Cl2
0.349 moles Cl2 reacts with 0.354 * 2/3 (= 0.236) moles of Al to produce same amount (0.236 moles) of AlCl3
THEREFORE some of the Al is left unreacted.
yield of product in grams = molar mass * no of moles = 133.5 * 0.236 = 31.562 g of AlCl3


Hope you get it.

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