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# A-Level Chemistry Question watch

1. I dont suppose anyone could help with this (really hard)

[/[/B]I] 10g of a mixture of copper powder and magnesium powder was mixed with 100cm³ of 1.0 mol dm³ hydrochloric acid. The copper does not react, but the magnesium does as shown:

Mg + 2HCL = MgCl² + h²

The Resulting solution was filtered to remove unreacted copper and then made up to 250cm³ with water. 25.0cm³ of this solution was found to neutralise 36.8cm³ of 0.2 mol dm³ NaOH. Find the % by mass of the magnesium in the metal powder mixture.[/[/B]I]

So far i have done this have no clue though if i am on the right tracks
0.2mold/0.025 = 8 moles of NaOH in 25cm³ therefore 80moles in 250cm³

can anyone help i dont know whether this is right and if it is what to do next

thank you very very much
2. OK, im not sure if this is right but I THINK iM ON THE RIGHT LINES

The final reaction is mgcl2 + 2NaOH = NaCl + Mg(OH)2

moles of NaOH = (36.8 x 0.2)/1000 = 7.36x10^-3

The ration of MgCl2 and NaOH is 1:2

Therefore there are 3.68x10^-3 moles of MgCl2 present

If 25cm^3 of MgCl2 has 3.68x10^-3 moles
the 250cm^3 has 0.0368 moles

Im pretty sure thats right upto there, but Im not quite sure where to go from here
3. ok, ive worked it out now:

now work out the moles of HCl that has been originally taken
so (100 x 1.0)/1000 = 0.1

Therefore the amount of HCl that has reacted = 0.1 - 0.0368 = 0.0632 moles

Using the equation given the ratio of Mg : HCl is 1:2

Therefore there are 0.5 x 0.0632 moles of Mg =0.0316

therefore mass of Mg is 0.0316 x 24 = 0.76g

Im pretty sure this is correct
4. Thanks alot, is that realistic a 10g are going in, i dont know. Thanks can i ask are you doing A-level or degree level chemistry?
5. I have just finished my A-levels (AAAa in chem, physics, maths and biology) and am currently studying medicine at newcastle, but am reapplying to study chem at oxford next year.
6. Oh right i have just started year 12 (As Chemistry) and am finding it quite challenging. Do you think you might be able to help me with this question i dont know anything about purity.

A solution is made up by dissolving 1.25g of impure sodium hydroxide in water and making it up to 250cm³ of solution. 25.0cm³ of this solution is neutralised by 30.3cm³ of 0.102 M HCl. Calculate the percentage purity of the sodium hydroxide.
7. OK, based on a pretty similar principle to that above:

You know (I hope) that an acid and a base give a salt and water

NaOH + HCl = NaCl + H2O

moles of HCl = (30.3 x 0.102)/1000 = 3.091x10^-3

The ratio of NaOH to HCl is 1:1

Therefore there are 3.091x10^-3 moles of NaOH in 25cm^3

Therefore in 250cm^3 there are 0.03091 moles of NaOH

Since moles = mass/Mr

mass of pure NaOH = moles x Mr

=0.03091 x 40
=1.2364g
8. Thanks can anyone check my working for the following question

4.29g of crystalline sodium carbonate (Na²CO³xH²0) was made up to 250cm³ of aqueous solution 25.00cm³ of this solution required 15.0cm³ of 0.2 mol dm³ HCL for neutralisation. Calculate x, the number of molecules of water associated with each Na²CO³ unit.

My working:

moles = conc * volume
0.2 * (15/1000) = 0.003 moles of HCL
Ratio of HCL : Na²CO³xH²O = 1:1 (unsure if this is the right ratio)
therefore 0.003 moles of Na²CO³xH²0 in 25cm³
mass/no of moles = mr
4.29/0.003 = 1430 mr of Na²CO³xH²O
Mr of Na²CO³ = 106
1430-106 = 1324
divide by mr of H²O = 1324/18 = 73.5
therefore x=73.5
To me that seems a bit on the large size have i made a mistake anywhere?
9. Yup, thats wrong. Heres the solution

moles of HCl = (15 x 0.2)/1000 = 3x10^-3

You dont know how many molecules of water are assocuiated with the carbonate, HOWEVER, you know that an acid and a carbonate form a salt, water and carbondioxide

Therefore

Na2CO3.xH2O + HCL = NaCl + ......

Since there are 2 Na's you will form 2NaCl, and sop therefore you need 2 CL's so there are 2HCl

Threfore fatio of carbonate to HCl is 1:2

Therefore there ar 1.5x10^-3 moles of carbonate in 25cm^3
so 0.015 moles in 250cm^m

since moles = mass/Mr

0.015 = 4.29/(106 + 18x)

rearange to find x, which is 10
10. Thanks for that i was just unsure of how to get the ratio but it makes sense could you have a look at the working of one last question please.

A student did an experiment to find what mass of calcium carbonate there was in an indigestion table. She Crushed up a tablet and added an excess of hydrochloric acid (25.0cm³ of 1.0 mol dm³ HCL). She then titrated the excess acid against 0.50 mol dm³ NaOH, requiring 30.6 cm³ of the NaOH. Calculate the mass of CaCO³ in the tablet.

This is what i did but am certain its wrong

CaCO³ + 2HCL = CaCL² + H²0 + CO²
0.5 * (30.6/1000) = 0.0153 moles of NaOH
1.00*(25/1000) = 0.025 moles of HCL
CaCO³CL ratio 1:2
0.025/2 = 0.0125
moles * mr = mass
0.0125 * (16*3+40+12) = 1.25g

If you could have a look at my working i would be very grateful
11. OK, if you just think logically about the question and follow the steps above youl get the ansewr

moles of NaOh = (30.6 x 0.5)/1000 = 0.0153

now HCl +NaOH = NaCl +H2O

The HCl is in excess so therefore all the NaOH will react and there will be some HCl left over

so moles of HCl = (25 x 1)/1000 = 0.025

Threfore the left over HCl = 0.025 - 0.0153 = 9.7x10^-3

Since CaCO3 + 2HCl = CaCl2 +H20 +CO2

ratio of Caco3 to HCl is 1:2

Threfore there are 4.85x10-3 moles of CaCO3

Mr of caco3 is 100

Therefore mass of Caco3 = 100 x 4.85x10^-3

mass = 0.485 grams
12. Thanks for that and good luck with your chemistry in the future you seem very good at it.
13. cheers

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