Sciencemadness Discussion Board

Unsure about this practice exam question.

chemic_oll - 25-4-2015 at 04:17

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.

chemic_oll - 25-4-2015 at 04:29

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.

Deathunter88 - 25-4-2015 at 05:28

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

blogfast25 - 25-4-2015 at 05:50

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.



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 %

chemic_oll - 26-4-2015 at 03:52

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 %


Thanks :) seems so simple once it's in front of me.