Chemistry - Mole concept (check) replied by secretliker @ Tue, 15 Jul 2008 17:26:21 +0800

secretliker — Tue, 15 Jul 2008 17:26:21 +0800

3rd question.

0.4mol of zinc atoms are added to 5cm3 (Is this 15cm3?) of 0.6mol/dm3 of silver ions.

=2.925g (Press calculator wrong! Should be 0.2925g if the above is 15cm3)

The others all correct.

Chemistry - Mole concept (check) replied by UltimaOnline @ Mon, 14 Jul 2008 21:41:22 +0800

UltimaOnline — Mon, 14 Jul 2008 21:41:22 +0800

If you're confident you know what you need to do for each of these questions (all of which are straightforward, nothing tricky), then the only reason why you might end up with the wrong answer, is due to careless mistakes.

Just to say (even though your post wasn't addressed to me, it was for everyone), I'm personally not inclined to work out all the questions to check your answers for you (or anyone else, for that matter). I'll only give advice and suggestions when asked for specific guidance or assistance on specific doubts, queries, concepts or topics.

However (don't get me wrong), it's still perfectly ok for you to make such posts here. I'm sure there are many students of your level (eg. 'O' or 'A') who would like the practice, and they can double check your answers for you.

Let's await replies from these other students, if any.

Chemistry - Mole concept (check) replied by bonkysleuth @ Mon, 14 Jul 2008 20:04:57 +0800

bonkysleuth — Mon, 14 Jul 2008 20:04:57 +0800

1. 3.2g of sivler nitrate was added to 50cm3 of 0.4mol/dm3 of calcium chloride. calculate the mass of silver chloride precipitated at the end of experiment.

The equation of reaction: 2AgNO3 + CaCl2 -> 2 AgCl + Ca(NO3)2

Mr of AgNO3 = 170

No of moles of AgNO3 = 3.2/170 = 0.01882352941 moles

50cm3 of CaCl2 = 0.05dm3

Concentration in mol/dm3 = no of moles of CaCl2 / vol. of soln.

0.4 = no of moles of CaCl2 / 0.05

.: no of moles = 0.4 *0.05 = 0.02 moles

Limiting factor is silver nitrate.

Compare silver nitrate with silver chloride. (in terms of mole ratio)

no of moles needed for Agcl = 0.1882352941 moles

Find Mr of AgCl = 143.5

Mass of AgCl: 143.5 *0.018823352941

= 2.70g

2. Aqueous barium chloride reacts with sulphuric acid according to equation below

BaCl2 + H2SO4 -> BaSO4 + 2HCl

a. calculate the mass of baiurm sulphate precipitate produced when 250cm3 of 0.2mol/dm3 of sulphuric acid reacts with 200cm3 of 0.25mol/dm3 of barium chloride.

Answer's too lengthy and I don't have much time to spare... so i'll just show part of the answer.

In this question, there's no limiting factor as no. of moles for the reactants are exactly the same.

The answer I got is 11.65g

3rd question.

zinc reacts with silver ions according to the equation : Zn + 2Ag+ ->Zn2+ + 2 Ag.

calculate mass of zincions formed when 0.4mol of zinc atoms are added to 5cm3 of 0.6mol/dm3 of silver ions.

limiting factor is Ag+ for this case.

Mole ratio for Ag+ : Zn2+

2 : 1

= 0.009 : 0.0045

Last part: is Mr of Zn2+ same as zinc atoms 65? if thats the case, then

0.0045 = mass of zn2+ / 65

mass = 65 *0.0045

=2.925g

Help me check the following questions. but they should be correct, as far as i know. find limiting factor for each case.

1. calcium hydroxide : 7.5g/dm3 and 20 cm3

Nitric acid : 0.81mol/dm3 and 25cm3

limiting factor is calcium hydroxide (ans)

2. Potassium hydroxide: 12g/dm3 and 32cm3

Ammomium sulphate - 2.9g

limiting factor is potassium hydroxide(ans)

3. calcium metal - 6g

sulphuric acid : 0.8mol in 500cm3

calcium has 0.15moles and suplhuric acid 0.8moles.

limiting factor is calcium. (ans)

for the last 3 questions, please try to work out the statements on your own because i'm in a rush for time. sorry... and thanks!

Recent Posts in 'Chemistry - Mole concept (check)' | sgForums.com

Chemistry - Mole concept (check) replied by secretliker @ Tue, 15 Jul 2008 17:26:21 +0800

Chemistry - Mole concept (check) replied by UltimaOnline @ Mon, 14 Jul 2008 21:41:22 +0800

Chemistry - Mole concept (check) replied by bonkysleuth @ Mon, 14 Jul 2008 20:04:57 +0800