# Displacement

Watch
Announcements
#1
CuSO4 Mg - Cu MgSO4Calculate the mass of magnesium needed to displace all of the copper from 100 cm3 of copper sulfate solutionConcentration of copper sulfate solution is 50 g CuSO4 per dm3
0
4 months ago
#2
(Original post by Manni8194)
CuSO4 Mg - Cu MgSO4Calculate the mass of magnesium needed to displace all of the copper from 100 cm3 of copper sulfate solutionConcentration of copper sulfate solution is 50 g CuSO4 per dm3
Star by working out the moles of copper sulfate in the solution.
Last edited by EierVonSatan; 4 months ago
0
#3
(Original post by EierVonSatan)
Star by working out the moles of copper sulfate in the solution.
Then what do you do after that?
0
3 months ago
#4
(Original post by Manni8194)
Then what do you do after that?
Use the balanced chemical equation to work out the number of moles of Mg needed to react.

Once you've got that it's easy to convert that to mass by using the Ar 0
#5
(Original post by EierVonSatan)
Use the balanced chemical equation to work out the number of moles of Mg needed to react.

Once you've got that it's easy to convert that to mass by using the Ar :yes
K thank you 😊
Last edited by Manni8194; 3 months ago
0
3 months ago
#6
(Original post by Manni8194)
K thank you 😊
Yep. What did you get?
0
#7
(Original post by EierVonSatan)
Yep. What did you get?
I tried to work it out but im confused on how to do it. I would really appreciate it if you showed me how to do it step by step. If not its okay thanks for ur help anyway. 0
3 months ago
#8
(Original post by Manni8194)
I tried to work it out but im confused on how to do it. I would really appreciate it if you showed me how to do it step by step. If not its okay thanks for ur help anyway. Instread of just giving you the answer, here is a similar question, done step by step (I am going to assume you are using a GCSE periodic table)

Mg + FeSO4 ---> MgSO4 + Fe
Calculate the mass of magnesium needed to displace all of the iron from 250 cm3 of iron (II) sulfate solution. The concentration of iron (II) sulfate solution is 12.0 g/dm3

Step 1: Working out the moles of iron sulfate present! If there are 12g in 1000cm3 (= 1 dm3), then there are 3g in 250cm3 (i.e. dividing by 4)
to convert from g to mol, we use the formula moles = mass/Mr = 3g/(56+32+4x16) = 3/152 = 0.0197...mol of iron (II) sulfate

Step 2: From the balanced equation above, for every one FeSO4 that gets reacted so does one Mg. Therefore, there is also going to be 0.0197...mol of magnesium needed to react all the FeSO4

Step 3: We now need to convert the moles into mass, using the above formula. Mass = moles x Ar = 0.0197... x 24 = 0.474g (3sf)
0
#9
(Original post by EierVonSatan)
Instread of just giving you the answer, here is a similar question, done step by step (I am going to assume you are using a GCSE periodic table)

Mg + FeSO4 ---> MgSO4 + Fe
Calculate the mass of magnesium needed to displace all of the iron from 250 cm3 of iron (II) sulfate solution. The concentration of iron (II) sulfate solution is 12.0 g/dm3

Step 1: Working out the moles of iron sulfate present! If there are 12g in 1000cm3 (= 1 dm3), then there are 3g in 250cm3 (i.e. dividing by 4)
to convert from g to mol, we use the formula moles = mass/Mr = 3g/(56+32+4x16) = 3/152 = 0.0197...mol of iron (II) sulfate

Step 2: From the balanced equation above, for every one FeSO4 that gets reacted so does one Mg. Therefore, there is also going to be 0.0197...mol of magnesium needed to react all the FeSO4

Step 3: We now need to convert the moles into mass, using the above formula. Mass = moles x Ar = 0.0197... x 24 = 0.474g (3sf)
Thank you soo much this helps a lot. I couldnt find this anywhere, thank u soooo much ur amazing 😊.And is the answer to my question 0.72?
Last edited by Manni8194; 3 months ago
0
3 months ago
#10
(Original post by Manni8194)
Thank you soo much this helps a lot. I couldnt find this anywhere, thank u soooo much ur amazing 😊.And is the answer to my question 0.72?
0.752g

moles of CuSO4 = 0.0313... mol

1:1 ratio so 0.0313...mol of Mg

Mass = 0.0313...x24 = 0.752g
0
#11
(Original post by EierVonSatan)
0.752g

moles of CuSO4 = 0.0313... mol

1:1 ratio so 0.0313...mol of Mg

Mass = 0.0313...x24 = 0.752g
Hello its me again sorry just wanted to say i followed ur example and technique and it helped, i know understand it so just wanted to say thank u sooo much for the help really appreciate it   1
3 months ago
#12
(Original post by Manni8194)
Hello its me again sorry just wanted to say i followed ur example and technique and it helped, i know understand it so just wanted to say thank u sooo much for the help really appreciate it   Very welcome!
1
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (525)
33.78%
Yes, I like the idea of applying to uni after I received my grades (PQA) (642)
41.31%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (315)
20.27%
I think there is a better option than the ones suggested (let us know in the thread!) (72)
4.63%