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