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Not sure if this is right bruv, haven't done these in a while, but here goes:
Find the moles of HCl = (VM)/1000 = (0.78 x 15.4)/1000 = 0.012012
Ratio of moles of HCl to moles of Na2CO3.XH2O = 2:1
So Moles of Na2CO3.XH2O = 0.012012/2 = 0.006006
0.006006 is the number of moles in 25 cm3, in 250 cm3 you have 10 times as much, so you multiply the value by 10 = 0.06006 moles.
Now some algebra time:
Using Mass = moles x Mr
Total mass = 17.16
Total Moles = 0.06006
Mr = 106 + 18X
Plug them into the formula above:
17.16 = 0.06006 x (106 + 18X)
17.16/0.06006 = 106 + 18X
(285.714 - 106)/18 = X
X = 9.9841
Answer looks reasonable, if I've gone wrong somewhere, feel free to point it out.
And thanks for the rep for the other question the other day.
EDIT: I got the Mr of of Na2CO3 wrong, didn't have a periodic table handy, should have been 106 instead of 116. This gives a final answer of 9.9841 or about 10 which looks even more reasonable than before.
Find the moles of HCl = (VM)/1000 = (0.78 x 15.4)/1000 = 0.012012
Ratio of moles of HCl to moles of Na2CO3.XH2O = 2:1
So Moles of Na2CO3.XH2O = 0.012012/2 = 0.006006
0.006006 is the number of moles in 25 cm3, in 250 cm3 you have 10 times as much, so you multiply the value by 10 = 0.06006 moles.
Now some algebra time:
Using Mass = moles x Mr
Total mass = 17.16
Total Moles = 0.06006
Mr = 106 + 18X
Plug them into the formula above:
17.16 = 0.06006 x (106 + 18X)
17.16/0.06006 = 106 + 18X
(285.714 - 106)/18 = X
X = 9.9841
Answer looks reasonable, if I've gone wrong somewhere, feel free to point it out.
And thanks for the rep for the other question the other day.
EDIT: I got the Mr of of Na2CO3 wrong, didn't have a periodic table handy, should have been 106 instead of 116. This gives a final answer of 9.9841 or about 10 which looks even more reasonable than before.
Question 7. First find the number of moles of NA2CO3 so thats n=mass(g)/mr
find number of moles of HCL using n=vm/1000. Not sure what you do after that sorry lets hope someone else can help.
Question 8. First find number of moles of Hydrogen Peroxide which is n=vm/1000. so thats 0.0004 moles. Its a 5:2 mole ratio so divide 0.0004 by 5 which is 0.00008 moles of hydrogen perxoide.
rearragne to find volume so thats v=n*1000/M = 0.00008*1000/0.04=2cm3 of hydrogen peroxide. Just make sure its right, it looks right to me. good luck mate
find number of moles of HCL using n=vm/1000. Not sure what you do after that sorry lets hope someone else can help.
Question 8. First find number of moles of Hydrogen Peroxide which is n=vm/1000. so thats 0.0004 moles. Its a 5:2 mole ratio so divide 0.0004 by 5 which is 0.00008 moles of hydrogen perxoide.
rearragne to find volume so thats v=n*1000/M = 0.00008*1000/0.04=2cm3 of hydrogen peroxide. Just make sure its right, it looks right to me. good luck mate
8a. Look at which reagent you have most information about, which is the peroxide. So find the number of moles of peroxide using n = (MV)/1000
Moles hydrogen peroxide: (40 x 0.1)/1000 = 0.0004
To find the number of moles of the manganate:
Its a 5:2 ratio so divide the number of moles by 5 and then multiply it by 2:
(0.0004/5) x 2 = 0.00016
Then find the volume of Manganate using V = (n x 1000)/M
V = 0.00016 x 1000 / 0.04 = 4 cm3
b. Moles of oxygen is the same as the moles of peroxide as they are both in a 5:5 ratio.
So all you do is multiply the moles of peroxide by 22.4
= 0.0004 x 22.4 = 0.0896 dm3
EDIT: To clarify for Neilh, 1 mole of any ideal gas occupies 22.4 dm3, I'm sure you already know that Zakz.
Moles hydrogen peroxide: (40 x 0.1)/1000 = 0.0004
To find the number of moles of the manganate:
Its a 5:2 ratio so divide the number of moles by 5 and then multiply it by 2:
(0.0004/5) x 2 = 0.00016
Then find the volume of Manganate using V = (n x 1000)/M
V = 0.00016 x 1000 / 0.04 = 4 cm3
b. Moles of oxygen is the same as the moles of peroxide as they are both in a 5:5 ratio.
So all you do is multiply the moles of peroxide by 22.4
= 0.0004 x 22.4 = 0.0896 dm3
EDIT: To clarify for Neilh, 1 mole of any ideal gas occupies 22.4 dm3, I'm sure you already know that Zakz.
17.16g of hydrated sodium carbonate of formula Na2CO3.XH2O were dissolved in water and made up to exactly 250cm^3 in a standard flask. 25cm^3 of this solution were then pipetted into a conical flask and titrated with 0.78M hydrochloric acid; 15.4cm^3 were required. Calculate the value of X in the formula.
Na2CO3.XH2O + 2HCL ----> 2NaCl + CO2 + (x+1)H2O
I would think the first step would be to calculate the amount of moles in the 25cm^3 solution pipetted into the conical flask. We can do this by working out the the amount of moles in the 250cm^3 solution. To do this we have to use the values given to us, we are given the mass and the RMM we can look at the periodic table and take its values. We can then work out the number of moles in the 250cm^3 solution by using the equation Moles = Mass/RMM.
So leaving us at,
Moles = 17.16g/(23+23+12+48+2+16) = 0.1384mol in 250cm^3 (0.25dm^3)
Now we can work out how many moles are in 25cm^3, we can do this by taking 250/25 = 10, by finding out how we can get from 250 to 25 we can apply the same principle to the moles,
0.1384/10 = 0.01384mol in 25cm^3
Okay I think I haven't been taught these types of questions yet , but I'll still give it a try.
Well what I suggest we do is find the mass of all the elements except H2O then you can find x+1 then once you found x+1 you can work out x by just -1 .
Lets start off by calculating the new mass of the Na2CO3.XH2O since we have 25cm^3 now.
So 0.01384mol * RMM will give us the new mass RMM being 124 will give us 1.716.
Now lets calculate the mass of HCl, we can do this by applying the ratio principle. The ratio of Na2CO3.XH20 : HCl is 1:2 which means I have to multiply the amount of moles by 2. Therefore 2 x 0.01384 = 0.02768. Now lets multiply the amount of moles by the RMM to calculate the moles.
0.02768 x 36.5 = 1.01032
Then we go through the same process to calculate the other masses. 1) Look at the ratios, 2) Multiply moles by ratio and then multiply by the RMM to calculate the mass of the compound.
Mass of 2NaCl: 1.61928
Mass of CO2: 0.60896
Now that we have all the masses we can work out the mass of the H2O,
I done this by,
1.716 + 1.01032 = 1.61928 + 0.60896 + Mass of H2O
Mass of h2O = 0.49808
0.4908/RMM = Moles of H2O = 0.027671111
Now divide the moles of H2O by the moles of a compound with a coefficient of one to get the (x+1) of the equation.
0.027671111/0.01384 = 1.9994 which I think we can round up to 2 since i didn't take the exact value in earlier calculations.
So x+1 = 2 therefore x = 1
Hmm hope that's correct...
Anyway, as usual, if anyone sees a flaw in my method please point it out. As for the other question if i have enough time later on I'll give it a go.
Take care
Hamid
Na2CO3.XH2O + 2HCL ----> 2NaCl + CO2 + (x+1)H2O
I would think the first step would be to calculate the amount of moles in the 25cm^3 solution pipetted into the conical flask. We can do this by working out the the amount of moles in the 250cm^3 solution. To do this we have to use the values given to us, we are given the mass and the RMM we can look at the periodic table and take its values. We can then work out the number of moles in the 250cm^3 solution by using the equation Moles = Mass/RMM.
So leaving us at,
Moles = 17.16g/(23+23+12+48+2+16) = 0.1384mol in 250cm^3 (0.25dm^3)
Now we can work out how many moles are in 25cm^3, we can do this by taking 250/25 = 10, by finding out how we can get from 250 to 25 we can apply the same principle to the moles,
0.1384/10 = 0.01384mol in 25cm^3
Okay I think I haven't been taught these types of questions yet
, but I'll still give it a try.Well what I suggest we do is find the mass of all the elements except H2O then you can find x+1 then once you found x+1 you can work out x by just -1 .
Lets start off by calculating the new mass of the Na2CO3.XH2O since we have 25cm^3 now.
So 0.01384mol * RMM will give us the new mass RMM being 124 will give us 1.716.
Now lets calculate the mass of HCl, we can do this by applying the ratio principle. The ratio of Na2CO3.XH20 : HCl is 1:2 which means I have to multiply the amount of moles by 2. Therefore 2 x 0.01384 = 0.02768. Now lets multiply the amount of moles by the RMM to calculate the moles.
0.02768 x 36.5 = 1.01032
Then we go through the same process to calculate the other masses. 1) Look at the ratios, 2) Multiply moles by ratio and then multiply by the RMM to calculate the mass of the compound.
Mass of 2NaCl: 1.61928
Mass of CO2: 0.60896
Now that we have all the masses we can work out the mass of the H2O,
I done this by,
1.716 + 1.01032 = 1.61928 + 0.60896 + Mass of H2O
Mass of h2O = 0.49808
0.4908/RMM = Moles of H2O = 0.027671111
Now divide the moles of H2O by the moles of a compound with a coefficient of one to get the (x+1) of the equation.
0.027671111/0.01384 = 1.9994 which I think we can round up to 2 since i didn't take the exact value in earlier calculations.
So x+1 = 2 therefore x = 1
Hmm hope that's correct...
Anyway, as usual, if anyone sees a flaw in my method please point it out. As for the other question if i have enough time later on I'll give it a go.
Take care
Hamid
JayEm
8a. Look at which reagent you have most information about, which is the peroxide. So find the number of moles of peroxide using n = (MV)/1000
Moles hydrogen peroxide: (40 x 0.1)/1000 = 0.0004
To find the number of moles of the manganate:
Its a 5:2 ratio so divide the number of moles by 5 and then multiply it by 2:
(0.0004/5) x 2 = 0.00016
Then find the volume of Manganate using V = (n x 1000)/M
V = 0.00016 x 1000 / 0.04 = 4 cm3
b. Moles of oxygen is the same as the moles of peroxide as they are both in a 5:5 ratio.
So all you do is multiply the moles of peroxide by 22.4
= 0.0004 x 22.4 = 0.0896 dm3
Moles hydrogen peroxide: (40 x 0.1)/1000 = 0.0004
To find the number of moles of the manganate:
Its a 5:2 ratio so divide the number of moles by 5 and then multiply it by 2:
(0.0004/5) x 2 = 0.00016
Then find the volume of Manganate using V = (n x 1000)/M
V = 0.00016 x 1000 / 0.04 = 4 cm3
b. Moles of oxygen is the same as the moles of peroxide as they are both in a 5:5 ratio.
So all you do is multiply the moles of peroxide by 22.4
= 0.0004 x 22.4 = 0.0896 dm3
where has that 22.4 come from jayem?
Hamid16
haha, it seems that my teacher made a typo in the question I wanted help in so never mind, sorry my answer wasn't correct zakz... 
Are you doing Chemistry with Edexcel?
Are you doing Chemistry with Edexcel?Yeah, your question isn't doable because you don't have a volume. 17.1 gdm-1 means 17.1 grams in 10cm which doesn't make sense.
If its was 17.1 gdm-3, you just divide 17.1 by Barium hydroxide's Mr to get the value in mol dm-3
Jay, our teacher mailed us the answer to the algebra Q:
Na2CO3.xH2O(s) + 2HCl(aq) —» 2NaCl(aq) + CO2(g) + (x+l)H20(l)
Moles of HCl = 15.4/1000 x 0.78 = 0.012
Ratio hydrochloric acid : sodium carbonate = 2:1
Moles of sodium carbonate in 25cm3 = 0.012/2 = 0.006
Moles of sodium carbonate in 250 cm3 = 0.006 x 10 = 0.06
Therefore 17.16g hydrated sodium carbonate = 0.06 moles
This means that the RFM of hydrated sodium carbonate = 0.06/17.16 = 286
RFM of anhydrous sodium carbonate (Na2CO3) = 106
Mass due to water = 286 – 106 = 180g
Moles of water = 180/18 = 10
Therefore x = 10
You are the man! lol ...psst u have more accuracy than him
Na2CO3.xH2O(s) + 2HCl(aq) —» 2NaCl(aq) + CO2(g) + (x+l)H20(l)
Moles of HCl = 15.4/1000 x 0.78 = 0.012
Ratio hydrochloric acid : sodium carbonate = 2:1
Moles of sodium carbonate in 25cm3 = 0.012/2 = 0.006
Moles of sodium carbonate in 250 cm3 = 0.006 x 10 = 0.06
Therefore 17.16g hydrated sodium carbonate = 0.06 moles
This means that the RFM of hydrated sodium carbonate = 0.06/17.16 = 286
RFM of anhydrous sodium carbonate (Na2CO3) = 106
Mass due to water = 286 – 106 = 180g
Moles of water = 180/18 = 10
Therefore x = 10
You are the man! lol
...psst u have more accuracy than him 
please can somebody help me with this?! http://www.thestudentroom.co.uk/showthread.php?p=41185631&highlight= similar (uses find x Na2CO3.xH2O) but when i do the methods you've done i end up with reaaaaly large answers x
I dont know if i got lucky but i got the answer this way and it was real quick.
I did 17.16 X 0.1 = 1.716 grams
as the original solution is 250 cm cubed and the new one has 25 cubed so 25/250 = 0.1
and we can times o.1 by 17.16 to find its new mass
the moles of HCL is 0.012012 mol (0.78 mol dm X 0.0154 Dm)
and the mole ratio is 1:2
so we divide it by 2 = 0.006006
and we can use the formula mass/moles = the Mr
so 1.716/0.006006 = 285.7
so now we know we can turn it into a equation kinda
mass of hydrated Na2CO3. x H20 = 105.9 +18x =285.7
x= 9.98 = 10
I did 17.16 X 0.1 = 1.716 grams
as the original solution is 250 cm cubed and the new one has 25 cubed so 25/250 = 0.1
and we can times o.1 by 17.16 to find its new mass
the moles of HCL is 0.012012 mol (0.78 mol dm X 0.0154 Dm)
and the mole ratio is 1:2
so we divide it by 2 = 0.006006
and we can use the formula mass/moles = the Mr
so 1.716/0.006006 = 285.7
so now we know we can turn it into a equation kinda
mass of hydrated Na2CO3. x H20 = 105.9 +18x =285.7
x= 9.98 = 10
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