# Chemistry a level question help!!

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1,2–dichloroethane undergoes thermal cracking to give chloroethene.

CH2ClCH2Cl ——> CH2=CHCl + HCl

Calculate the percentage yield of this process if 10.0 tonnes of the 1,2–dichloroethane yield 2.0 tonnes of chloroethene.

I worked this out and I got 40% but the correct answer is apparently 31.7% and I don’t understand why (they didn’t show working out). Can anyone please explain??

CH2ClCH2Cl ——> CH2=CHCl + HCl

Calculate the percentage yield of this process if 10.0 tonnes of the 1,2–dichloroethane yield 2.0 tonnes of chloroethene.

I worked this out and I got 40% but the correct answer is apparently 31.7% and I don’t understand why (they didn’t show working out). Can anyone please explain??

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#2

Start off by working out the number of moles of 1,2-dichloroethane that decomposes. Moles (mol) = mass (g) / Mr so the moles will be 10,000,000/99 which is 101,010.101 moles.

From the equation we can see that one mole of 1,2-dichloroethane produces one mole of chloroethene, so there will also be 101,010.101 moles of chloroethene. We can use the equation again to work out the mass of chloroethene this should have produced: 101010.101 x 62.5 = 6,313,131.313 g or 6.313 tonnes.

Then divide the actual mass produced by the theoretical mass produced and multiply by 100 to get the percentage yield: (2 / 6.313) x 100 = 31.7%. Hope this helps .

From the equation we can see that one mole of 1,2-dichloroethane produces one mole of chloroethene, so there will also be 101,010.101 moles of chloroethene. We can use the equation again to work out the mass of chloroethene this should have produced: 101010.101 x 62.5 = 6,313,131.313 g or 6.313 tonnes.

Then divide the actual mass produced by the theoretical mass produced and multiply by 100 to get the percentage yield: (2 / 6.313) x 100 = 31.7%. Hope this helps .

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(Original post by

Start off by working out the number of moles of 1,2-dichloroethane that decomposes. Moles (mol) = mass (g) / Mr so the moles will be 10,000,000/99 which is 101,010.101 moles.

From the equation we can see that one mole of 1,2-dichloroethane produces one mole of chloroethene, so there will also be 101,010.101 moles of chloroethene. We can use the equation again to work out the mass of chloroethene this should have produced: 101010.101 x 62.5 = 6,313,131.313 g or 6.313 tonnes.

Then divide the actual mass produced by the theoretical mass produced and multiply by 100 to get the percentage yield: (2 / 6.313) x 100 = 31.7%. Hope this helps .

**Thiexp53**)Start off by working out the number of moles of 1,2-dichloroethane that decomposes. Moles (mol) = mass (g) / Mr so the moles will be 10,000,000/99 which is 101,010.101 moles.

From the equation we can see that one mole of 1,2-dichloroethane produces one mole of chloroethene, so there will also be 101,010.101 moles of chloroethene. We can use the equation again to work out the mass of chloroethene this should have produced: 101010.101 x 62.5 = 6,313,131.313 g or 6.313 tonnes.

Then divide the actual mass produced by the theoretical mass produced and multiply by 100 to get the percentage yield: (2 / 6.313) x 100 = 31.7%. Hope this helps .

I’m also struggling with the next question.

Use the percentage yield and atom economy of this reaction to calculate how much in tonnes of the 1,2–dichloroethane is actually converted into chloroethene.

Any help would be appreciated

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#4

Percentage yield we know from the previous question (31.7%). Atom economy we calculate by dividing the Mr of the desired product by the Mr of all reactants and multiplying by 100, in this case it would be (62.5 / 99) x 100 which equals 63.1%.

Then to work out how much of the 1,2-dichloroethane turns to chloroethene we multiply the mass of 1,2-dichloroethane, the percentage atom economy and the percentage yield together: 0.317 x 0.631 x 10 = 2.00126 tonnes (or just 2.00 tonnes to 3 s.f.).

Then to work out how much of the 1,2-dichloroethane turns to chloroethene we multiply the mass of 1,2-dichloroethane, the percentage atom economy and the percentage yield together: 0.317 x 0.631 x 10 = 2.00126 tonnes (or just 2.00 tonnes to 3 s.f.).

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(Original post by

Percentage yield we know from the previous question (31.7%). Atom economy we calculate by dividing the Mr of the desired product by the Mr of all reactants and multiplying by 100, in this case it would be (62.5 / 99) x 100 which equals 63.1%.

Then to work out how much of the 1,2-dichloroethane turns to chloroethene we multiply the mass of 1,2-dichloroethane, the percentage atom economy and the percentage yield together: 0.317 x 0.631 x 10 = 2.00126 tonnes (or just 2.00 tonnes to 3 s.f.).

**Thiexp53**)Percentage yield we know from the previous question (31.7%). Atom economy we calculate by dividing the Mr of the desired product by the Mr of all reactants and multiplying by 100, in this case it would be (62.5 / 99) x 100 which equals 63.1%.

Then to work out how much of the 1,2-dichloroethane turns to chloroethene we multiply the mass of 1,2-dichloroethane, the percentage atom economy and the percentage yield together: 0.317 x 0.631 x 10 = 2.00126 tonnes (or just 2.00 tonnes to 3 s.f.).

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#6

**Thiexp53**)

Start off by working out the number of moles of 1,2-dichloroethane that decomposes. Moles (mol) = mass (g) / Mr so the moles will be 10,000,000/99 which is 101,010.101 moles.

From the equation we can see that one mole of 1,2-dichloroethane produces one mole of chloroethene, so there will also be 101,010.101 moles of chloroethene. We can use the equation again to work out the mass of chloroethene this should have produced: 101010.101 x 62.5 = 6,313,131.313 g or 6.313 tonnes.

Then divide the actual mass produced by the theoretical mass produced and multiply by 100 to get the percentage yield: (2 / 6.313) x 100 = 31.7%. Hope this helps .

**Thiexp53**)

Percentage yield we know from the previous question (31.7%). Atom economy we calculate by dividing the Mr of the desired product by the Mr of all reactants and multiplying by 100, in this case it would be (62.5 / 99) x 100 which equals 63.1%.

Then to work out how much of the 1,2-dichloroethane turns to chloroethene we multiply the mass of 1,2-dichloroethane, the percentage atom economy and the percentage yield together: 0.317 x 0.631 x 10 = 2.00126 tonnes (or just 2.00 tonnes to 3 s.f.).

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