11Q-1 Detonating gas is a mixture of oxygen and hydrogen gases . Amount of hydrogen and oxygen in the mixture is such that
volume of hydrogen in the mixturevolume of oxygen in the mixture=2
Q-2 let the total mass of mixture of MgCo3 and CaCo3 =100gm
now let mass of MgCo3 in the mixture be =x gm
..............mass of CaCo3 = (100-x) gm
When the given compunds will be heated following reaction will take place:
MgCo3 → MgO + Co2
moles→ x84 x84 x84
CaCo3 → Cao + Co2
moles→ (100-x)100 (100-x)100 (100-x)100
the loss in weight is due to escaping of Co2 from the mixture as a gas
therefore amount of Co2 escaped as gas = 50%of 100 gm = 50 gm
according to above chemical equations moles of Co2 formed = x84+ (100-x)100=k(say)
therefore k*(molar mass of Co2)=50
all the equations are given solve for x
%of MgCo3= 71.6%
% of CaCo3 = 28.4 %
11
vaibhav sharma
·2010-06-29 09:14:33
Q-3 I dont really undersatnd the question are they reacting VO with Fe2o3 to form V2O5 ??
11@ Vaibhav, yeah they are reacting VO with Fe2O3.
The equation is
VO + Fe2O3---------------------> FeO + V2O5
49reaction is..
2VO + 3Fe2O3 → 6FeO + 2V2O5
molec wt of VO= 67
MW of Fe2O3= 160
MW of V2O5= 182
let weight of V2O5 be x
then we can see 1 mole of V2O5 is formed from 2 moles of VO and 3 moles of Fe2O3
thus x182 = min{22X67,5.753X160}
49
Subhomoy Bakshi
·2010-06-29 12:13:50
4) 3Hg + 2I2 → Hg2I2 + HgI2
thus whatever the amount of reactants, the molar ratio of the products will be 1:1
thus ratio of the prducts= 658:454(ratio of molecular weights!!)
11
vaibhav sharma
·2010-06-29 18:45:32
@subhomoy
hey there is no reaction as such:
3Hg + 2I2 → HgI2 + Hg2I2
i think Hg2I2 and HgI2 are formed in two seprate reactions
Hg + I2 → HgI2
2Hg + I2 → Hg2I2
but i maybe wrong
@khyati
plz check the answer
13Lekin, there is no harm in adding up ur two equations to get a compact third one...It is upto us how we express the reaction(s).
11
Khyati
·2010-06-30 04:45:12
answer to the 4th 0.532:1
11
Khyati
·2010-06-30 04:45:17
answer to the 4th 0.532:1
11
Khyati
·2010-06-30 04:45:49
answer to the 4th 0.532:1
11
vaibhav sharma
·2010-06-30 12:46:37
@avik
nonononono
the answer will come different if you add the equations
if you add the equations the answer will be what subhomoy wrote but that is not the answer
11
vaibhav sharma
·2010-06-30 13:47:26
if you add the equations the question becomes very easy ....i don;'t think it is that easy.
11
vaibhav sharma
·2010-07-03 20:21:38
lets take x gms of Hg and x gms of I2 present in the initial mixture
Hg + I2 → HgI2
2Hg + I2 → Hg2I2
let say that y gm of I2 reacts with 200254y gm Hg in the first reaction to produce 454254y gm of HgI2
then (x-y) gm of I2 reacts with 400254(x-y) gm Hg in the second reaction to produce 654254(x-y) gm of Hg2I2
as the total amount of Hg present in the mixture =x gm
therefore
200254y + 400254(x-y) = x
from here we get: 73x = 100 y
as discussed above ratio of weight of HgI2 and Hg2I2 is:
454/254y654/254(x-y)
now as you know ratio x:y so you can calculate the above required ratio
answer is comin-→ 0.532793:1
