Moles Again...

[nid404]

This is my original solution hpe u understand it.
i think u may
find it in attachment

OKK..FIRST OF ALL

I DIDNT GET Y U APPLIED RATIO PROPORTION WITH 168:9.3::18:x

y 18?
they have talked abt CO2 not H2O

ok..heres the sol.--------------------------

take only NAHCO3 and CO2 into consideration

168g NAHCO3 - 44g CO2 (by molar masseS)

this means that pure NAHCO3 168g will give out 44g of CO2

however here..2.24g of CO2 is given out ..

by ratio and proportion find out how much grams of NAHCO3 should be used to give out 2.24g of CO2

44g - 168g
2.24 - 8.55g

but...here 9.3g of NAHCO3 impure is used

this means that there is 9.3 - 8.55g of impurity = 0.75g of impurity

they've asked to calculate % purity

so what v do is

=> 8.55/9.3 * 100 = 92%

what do u think slvri and nid404?

i solved the question too rite nw and i got da same ans as u adi....92%

told him..

what mistake did u find in his sol.
i didnt understand the ratio proportion thingy.

shrey...answer check kar..

firstly he used the Mr of water instead of carbon dioxide and secondly 168 should be related to 44 (not 9.3) while x (mass of nahco3) should be related to 2.24
so....168->44
and..x->2.24
cross multipying we find x=8.55 g
then % purity = (8.55/9.3)*100=92% (approx)

I checked it too...adi and slvri are right

[@d!_†oX!©]

2NaHCO3 ---> Na2CO3 + H2O + CO2
Moles of NaHCO3 = 9.3/84 =0.1107 moles
Mole Ratio ---> 2:1
Thus, Moles of CO2 formed(theoretically) = 0.1107/2 =0.055 moles
And, Moles of CO2 formed(practically) = 2.24/44 = 0.0509 moles
Thus % Purity = (0.0509/0.055)*100
                   = 91.96% = 92%

[nid404]

yup it's long confirmed but thanks nyways

[@d!_†oX!©]

i know its confirmed nid.....i was just checking myself.....