IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Sciences => Topic started by: ~~~~shreyapril~~~~ on September 10, 2009, 02:16:59 pm
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I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!
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I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!
molar mass or CO2 is 12+2*16=44 so 2.2/44=0.05 moles of CO2 so .05 moles of NAHCO3. 1 mol of NaHCO3 weighs 23+1+60=84 and 0.5 moles weighs 0.05*84=4.2g
Purity =4.2/9.3=4.5=45%
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oops sorry
i typed the question wrong again..
it is 2.24g not 2.2
sry a silly mistake.
cant understand my own handwritting..
but thanx for the steps astar.
they will help me find the ans
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oops sorry
i typed the question wrong again..
it is 2.24g not 2.2
sry a silly mistake.
cant understand my own handwritting..
but thanx for the steps astar.
they will help me find the ans
doesnt change the answer much
46%
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i got some wierd ans wen i tried it the way a star told me
i got something near to 98.28% purity
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I am sure abt his question as my teacher had given me this one it is not like the previous one..
Here it starts
>>>>>>>>>>>>>>>>
9.3g of impure Sodium bicarbonate[NaHCO3] was heated and 2.2g of CO2 was formed.
Calculate % purity of NAHCO3
pls hepl!!!
a simpler way of diong this is use ratio proportion
44g of CO2 is produced with - 84g of pure NAHCO3
therefore 2.24g of CO2 should be produced with - 84*2/44 * 2.24 = 8.55
8.55 / 9.2 * 100 = 93%
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oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
k..
so acc to u it will be
168/44*2.24=8.55
so purity wud be 8.55/9.3*100=91.96 [app]
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oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation
astar missed it too..
u just rote CO2
yes its 84*2
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oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation
astar missed it too..
u just rote CO2
yes its 84*2
i'm getting 92%
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oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
k..
so acc to u it will be
168/44*2.24=8.55
so purity wud be 8.55/9.3*100=91.96 [app]
yes
thats wat i rote
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i got some wierd ans wen i tried it the way a star told me
i got something near to 98.28% purity
how ddid u get
98%
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ohh
but i think u had written 46 not 92 if i am not wrong..
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oye it shud be 84*2
the equation for the same is..
2NaHCO3 --> Na2CO3 + CO2 + H2O
oh..missed that..
u shud have ritten the whole equation
astar missed it too..
u just rote CO2
yes its 84*2
i'm getting 92%
chk this
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Mr of NAHCO3 is 168 Mr of CO2 is 44
Mass of NAHCO3 is 9.3 Mass of Na2CO3 is 5.9 mass of CO2 is 2.24 mass of H2O is 1
5.9+1+2.24=9.14
9.14/9.3*100=98.28[app]
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i THOUGHT THE EQUATION WOULD BE NaOH3=NaOH+CO2
Which make the ratio 1:1
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NaHCO3 has molar mass 23+1+12+3*16=84
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NaHCO3 has molar mass 23+1+12+3*16=84
sir the equation hs 2NaHCO3
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This is my original solution hpe u understand it.
i think u may
find it in attachment
-
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%
-
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?
-
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%
-
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..:)
-
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)
-
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)
true..
thanks for confirmation!
:)
-
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)
true..
thanks for confirmation!
:)
dont mention it ;) :D
and shrey......when r u giving igcse?
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we r giving it together...oct/nov
hez in my class.. :)
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we r giving it together...oct/nov
hez in my class.. :)
coooool........so wut r ur subs adi?
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math
phy
chem
eng
computer studies
french
add. math
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math
phy
chem
eng
computer studies
french
add. math
nice.....u take french too?
my mom knws both french and german but i havent learned either even though i have books in ma house of them.......
so is french hard? or easy?
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haha ;D
a silly mistake and the whole question went wrong
tks slvri and aadi
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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
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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%
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yup it's long confirmed but thanks nyways
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i know its confirmed nid.....i was just checking myself..... :) :)