Question

The purity os sodium hydrogen carbonate sample can be found bye experiment. In such a exercise, Mona added a carefully weighed NaHCO3 sample to a conical flask containing a concentrated hydrochloric solution. the data recorded is shown here:

most of impure NaHCO3...... 4.262g
Mass of flask + HCL solution ... 57.982g
mass of flask + contents after the reaction was complete.........60.370g

find the mass of NaHCO3 actually present in the impure measured sample.

Answer 1

@chmvijay

Answer 2

u guys might think im stupid but at this time - midnight- especially after doing maths for 5 hrs- i have nothing in my brain left. so guys please co-operate here

Answer 3

@RANE by the way who is mona

Answer 4

i dont know its just what the question states, she is my class mate to be honest but the teacher choose to use her name zoo

Answer 5

balanced equation ?

Answer 6

mass of flask + contents after the reaction was complete.........60.370g
is that including the Gas CO2 liberated in the reaction or what ?

Answer 7

this is equation but its no balancing

NaHCO3 + HCl = H20 + CO2 + NaCl

Answer 8

wait its already balanced

Answer 9

60.370g
is that including the Gas CO2 liberated in the reaction or what ?

Answer 10

idk this is all the information given, which i have posted above

Answer 11

i consider that they have weighed CO2 also and will try to solve it OK

Answer 12

most of impure NaHCO3+Mass of flask + HCL solution = 57.982g +4.262g =62.244 gram
but the total mass of the product is = 60.370g
so for 62.224 gram we are getting = 60.370 g
for 100 gram = 60.370 *100 /62.244 = claculate that is ur purity

Answer 13

vijay i did completely different to what u did

Answer 14

hmmm m fine ... thanks .. wt about u

Answer 15

what u did let me see that

Answer 16

iam fine musakan rane told me that ur her good friend

Answer 17

yea she is my best frend

Answer 18

n(NaHC03) = 0.045mols

n(NaHCO3) -actual [resent = 1/1*0.045mols

mass = 0.0045*9
= 4.0495g

Answer 19

good :)

Answer 20

lolz muskan ab hasti na rehna pls

Answer 21

hahha .. :D

Answer 22

according to ur answer its 95 % pure and from my answer its 97 % pure
:)
not much difference:)
how you got 0.045mols

Answer 23

nice smile @Muskan

Answer 24

tabhi nam muskan rakha hai .... :) q @RANE

Answer 25

n = m/M
= 4.262/91
= 0.045moles

Answer 26

hahah very smart @Muskan lakin mairy liyai to tum khuch aur hi ho

Answer 27

ahahha ... may tumhare liye tumhari DARLING hun u 9 :D

Answer 28

hahahah u never know

Answer 29

hahah .. chalo bata do kia hun tumhare liye may

Answer 30

i dont know really rane ur on other end of my calculation whcih i hat e a lot that is moles :P

Answer 31

lol i never knew u hated something in chemistry SHOCKING !!!

Answer 32

i also hate NaHCo3, H2SO4, CL2 etc.................................

Answer 33

thats the ones i love but atm i hate everything just wana go to bed and sleep

Answer 34

aur tumhe pata hai agy chemical eng karni hai may :_D

Answer 35

inshaalah lolz

Answer 36

hahaha tab tk sab se piayar ho jae ga

Answer 37

ha! we'll se "sab sy" !!!!

Answer 38

tum se to pehly din se hai ... baki sab means NaHCO3, CO2, H2O.....etc

Answer 39

hahaha

Answer 40

n(NaHC03) = 0.045mols

n(NaHCO3) -actual [resent = 1/1*0.045mols

mass = 0.0045*9
= 4.0495g
how u got this ?

Answer 41

:)