chemic_oll
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Unsure about this practice exam question.
I've started learning chemistry from home and have been trying a few English exam papers to see how I'm getting on. I'm not sure if I'm just being
daft and not understanding the question or if it's asking something I've not yet covered.
I believe the top question is correct but I'm a little confused by the next one. I understand relative formula mass and how to calculate the
percentage of an element within a compound as well as empirical formulas.
If anyone could point me in the right direction as to how I'd go about getting the answer that would be great.
I'd be grateful if the answer isn't given so I've chance to work things out myself.
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chemic_oll
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I'm having trouble uploading the pic from my phone.
1) iron(iii) chloride can be produced by the following reaction
2Fe + 3Cl2 - > 2FeCl3
Camculate the mass of iron chloride that can be produced from 11.20g of iron.
I answered 21.8g.
2) the actual mass of iron chloride produced was 24.3g. Calculate the % yield.
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Deathunter88
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Here is a really useful reaction balancing tool. It even has an option to set a limiting reagent and calculate how much of the other reagents are
needed and how much product will be produced.
http://theodoregray.com/PeriodicTable/MSP/BalanceReactions
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blogfast25
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Quote: Originally posted by chemic_oll  | I'm having trouble uploading the pic from my phone.
1) iron(iii) chloride can be produced by the following reaction
2Fe + 3Cl2 - > 2FeCl3
Camculate the mass of iron chloride that can be produced from 11.20g of iron.
I answered 21.8g.
2) the actual mass of iron chloride produced was 24.3g. Calculate the % yield.
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The molar mass of iron is 55.85 g/mol. 11.20 g Fe is thus 11.20/55.85 = 0.2005 mol. Complete conversion would yield 0.2005 mol of FeCl3. FeCl3 has a
molar mass of 162.2 g/mol, so 0.2005 mol is 0.2005 x 162.2 = 32.53 g FeCl3.
The percentage yield obtained was 24.3/32.53 x 100 % = 74.7 %
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chemic_oll
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| Quote: Originally posted by blogfast25 |
The molar mass of iron is 55.85 g/mol. 11.20 g Fe is thus 11.20/55.85 = 0.2005 mol. Complete conversion would yield 0.2005 mol of FeCl3. FeCl3 has a
molar mass of 162.2 g/mol, so 0.2005 mol is 0.2005 x 162.2 = 32.53 g FeCl3.
The percentage yield obtained was 24.3/32.53 x 100 % = 74.7 %
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Thanks  seems so simple once it's in front of me.
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