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# AQA chemistry help! Watch

1. (Original post by TobeTheHero)
I'm still confused?! do i use the answer to b) to work out how much hcl is present in 250?
but i worked how much hcl is present in 21.7cm^3 NOT 25?!

You were given the volume (21.7cm3) of NaOH, not HCl

You then found the moles of NaOH

From this you got the equivalent MOLES of HCl

etc
2. Thank You Guys! I worked out

I Understand Now! I'll post working soon
3. (Original post by TobeTheHero)
Question:

The chloride of an element Z reacts with water according to the following equation.

ZCL(4) (l) + 2H(2)O(l) > ZO(2) (s) + 4HCL(aq)

A 1.304g sample of ZCL(4) was added to water.

The solid ZO(2) was removed by filtration and the resulting solution was made up to 250cm(cubed) in a volumentric flask. A 25.0cm(cubed) portion of this solution
was titrated against a 0.112 M(molarity) of Sodium Hydroxide, of which 21.7cm(cubed) were required to reach the end point.

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Use this information to calculate the number of moles of HCL produced and hence the number of moles of ZCL(4) present in the sample.

Calculate the relative molecular masss, Mr, of ZCL(4).

From your answer deduce the relative atomic mass of, Ar, of element Z and hence its identity.

I've tried everything and i keep getting astronomical number for the Mr of ZCL(4)

1) HCl + NaOH > NaCl + Water

2) Moles of NaOH = 0.112 x 21.7/1000 = 0.0024304

3) Moles of HCl = moles of NaOH ratio 1:1

4) If 25 cm cubed contains 0.0024304 moles of HCl, then 250cm cubed contains 0.024304 mol

5) ratio is 1:4 so Moles of ZCl = 0.024304/4= 0.006075 mol

6) Mr of ZCl = 1.304/0.006075 = 214.6(5)

7) Ar of Z = 214.6(5)-(35.5 x 4) = 72.6(5)

8) Identity = ????
4. (Original post by TobeTheHero)
1) HCl + NaOH > NaCl + Water

2) Moles of NaOH = 0.112 x 21.7/1000 = 0.0024304

3) Moles of HCl = moles of NaOH ratio 1:1

4) If 25 cm cubed contains 0.0024304 moles of HCl, then 250cm cubed contains 0.024304 mol

5) ratio is 1:4 so Moles of ZCl = 0.024304/4= 0.006075 mol

6) Mr of ZCl = 1.304/0.006075 = 214.6(5)

7) Ar of Z = 214.6(5)-(35.5 x 4) = 72.6(5)

8) Identity = ????
This is where you pick up the periodic table and look for the element with a relative mass of 72.6

You would also hope that it has a valency of four...
5. (Original post by TobeTheHero)
1) HCl + NaOH > NaCl + Water

2) Moles of NaOH = 0.112 x 21.7/1000 = 0.0024304

3) Moles of HCl = moles of NaOH ratio 1:1

4) If 25 cm cubed contains 0.0024304 moles of HCl, then 250cm cubed contains 0.024304 mol

5) ratio is 1:4 so Moles of ZCl = 0.024304/4= 0.006075 mol

6) Mr of ZCl = 1.304/0.006075 = 214.6(5)

7) Ar of Z = 214.6(5)-(35.5 x 4) = 72.6(5)

8) Identity = ????
Much better! And the last step is the easiest one of all as charco says.

I must say, i really like that question. Really pushed you to understand molarity, stoichiometry and moles. Did your teacher write it or is it a textbook one?
Much better! And the last step is the easiest one of all as charco says.

I must say, i really like that question. Really pushed you to understand molarity, stoichiometry and moles. Did your teacher write it or is it a textbook one?
Teacher

At first this question baffled me, but once I carefully read the question again It started to come together and finally understood what it was asking for

7. HCl+NaOH---->NaCl+H20
moles of NaOH=0.112*(21.7/1000)
moles of NaOH=2.42*10^-3 this in 25cm^3
moles of NaOH=0.0243 this is in 250cm^3

HCl:ZCl4
4:1 molar ratio
0.0243/4=6.075*10^-3

Ar=m/moles Ar=1.304/6.075*10^-3=214.7gmol-1
Z=214.7-(35.5*4)=72.7gmol-1
Z=germanium
8. (Original post by helper1232)
HCl+NaOH---->NaCl+H20
moles of NaOH=0.112*(21.7/1000)
moles of NaOH=2.42*10^-3 this in 25cm^3
moles of NaOH=0.0243 this is in 250cm^3

HCl:ZCl4
4:1 molar ratio
0.0243/4=6.075*10^-3

Ar=m/moles Ar=1.304/6.075*10^-3=214.7gmol-1
Z=214.7-(35.5*4)=72.7gmol-1
Z=germanium
Why are you posting the answer to a question that is about 4 years old?

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