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 of Hg2I2 and HgI2 formed.(Hg=200 , I=127)

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 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 -→ 0.532793:1

 

2hg+i2=hg2i2,means 2x moles of hg react with one moles of i2 and gives one mole of hg2i2. 2x+x=x.hg+i2=hgi2, means one mole of hg react with one mole of i2 to gives one mole of hgi2. Y+y=y. According to question wt of hg and iodine are equal. So 200(2x+y)=254(x+y).from this eqton find x in term of y. Then put the value of x in this eqation 654x/454y which is the ratio of masses of hg2i2 and hgi2. Therefore we will get the ratio... Thnksss