Question

two metallic oxides contain 27.6%and30% oxygen respectively.if the formula of first oxide is X3O4. what is the formula for second one​

Answer 1

Answer:

X2O3 (or Fe2O3)

Explanation:

first metallic oxides contain 27.6 gram of O2 out of 100 gram

therefore mass of metal present is. 100 - 27.6 is equal to 72.4 gram

so So ratio of (moles of x / moles of O2) = 3/4

(72.4 ÷x) ÷(27.6 ÷ 16) = 3/4

on solving x = 55.9 = 56(Fe)

now for second metallic oxide contain 30 gram of O2 in hundred gram of compound therefore the mass of metal is 70 gram

so emperical formula = (moles of x /moles of O2) = (70÷56)÷(30÷16) =

(1.25/1.875) = 2/3

So , formula of second metallic oxide is

X2O3 (or Fe2O3).