Question

what is the percentage of oxygen in Fe2(SO4)3​

Answer 1

$\huge\underline{\underline{\sf Solution:}}$

Molecular Weight of $\sf{Fe_2(SO_4)_3}$

$\implies{\sf 2×56+3×32+12×16 }$

$\large\implies{\sf 400 g}$

1 mole of $\sf{Fe_2(SO_4)_3}$ contains 12 moles of Oxygen.

•°• 400g of $\sf{Fe_2(SO_4)_3}$ contains = 12 × 16

$\implies{\sf 192g\: Oxygen }$

Now ,

Percentage of Oxygen present is :

$\large\implies{\sf \dfrac{192}{400}×100 }$

$\large\implies{\sf 48 }$%

$\Large\underline{\underline{\sf Answer:}}$

•°• $\sf{Fe_2(SO_4)_3}$ contains 48% of Oxygen .

Answer 2

Answer

48 %

$\rule{100}2$

Explanation

To find the percentage of oxygen in ferric sulphate, we need to find molar mass of it. Then the percentage can be found.

Molar mass of $\sf Fe_2(SO_4)_3$

= 56 × 2 + (32 + 16 ×4) × 3

(since, molar mass of Fe = 56 g , S = 32 g and O = 16 g)

= 400 g

Now, total molar mass of O in $\sf Fe_2(SO_4)_3$

= 16 × 4 × 3

= 192 g

So, percentage of oxygen in ferric sulphate will be

⇒ 192/400 × 100

⇒ 192/4

= 48 %