1) 4.9g of KClO3, on heating shows a wt loss of 0.384g. What per% of the original KClO3 has decomposed?
2) 1.84g of of a mixture of CaCO3 and MgcO3 was heated to a constant weight. The constant weight of the residue was found to be 0.96g. Calculate the per% composition of the mixture.
3) Equal weights of mercury and iodine are allowed to react completely to form a mixture of mercurous and mercuric iodide leaving none of the reactants. Calculate the ratio by weight og Hg2I2 and HgI2 formed.
31..2KClO3→→2KCl +3O2
so 2 moles of KClO3 gives 3 moles of O2
so 1 mole of O2 is given by 2/3 moles of KClO3
now 0.384g. of O2 =0.012 moles of O2
again for this required KClO3 =2/3*0.012=0.008
but initially 4.9g of KClO3=0.04 moles of 4.9g of KClO3
so % decomposed =0.04-0.0080.04*100=80%
32..CaCO3→→CaO + CO2
x g
MgCO3→→MgO + CO2
(1.84-x)g
the residue is CaO and MgO =0.96g
as 1 mole CaCO3 gives 1 mole of CaO and 1 mole MgCO3 gives 1 mole of Mgo
so x g of CaCO3 =x/100 moles and that will give x/100*56 moles of CaO
similarily for MgCO3
and equate and get the value of x
x/100*56+(1.84-x)/84*40=0.96
and then find per% composition of the mixture.
11lets 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 200
254y gm Hg in the first reaction to produce 454
254y gm of HgI2
then (x-y) gm of I2 reacts with 400
254(x-y) gm Hg in the second reaction to produce 654
254(x-y) gm of Hg2I2
as the total amount of Hg present in the mixture =x gm
therefore
200
254y + 400
254(x-y) = x
from here we get: 73x = 100 y
as discussed above ratio of weight of HgI2 and Hg2I2 is:
454/254y
654/254(x-y)
now as you know ratio x:y so you can calculate the above required ratio
answer is comin-→ 0.532793:1
By 454/254y
654/254(x-y)
now as you know ratio x:y so you can calculate the above required ratio
answer is comin-→ 0.532793:1
By Vaibhav, since he solved this same question for me