Question

What is the mass of oxygen in 25.0 grams of potassium permanganate , KMnO4?

Answer 1

Answer: 10.1 g

I just did it and got i right

Answer 2

Answer:

10.76 grams

Explanation:

Given that the amount of $KMnO_4$ is 25.0 grams.

Mass of 1 mole of $KMnO_4$ = 158 grams

The number of atoms in 1 mole of $KMnO_4$ is 4.

Mass of oxygen in 1 mole of $KMnO_4$ = 16\times 4 = 68 grams.

Here, 158 grams of $KMnO_4$ has 68 grams of oxygen

So, the amount of oxygen in 1 gram of $KMnO_4$ = 68/158 grams

Therefore,  the amount of oxygen in 1 gram of $KMnO_4$ = $\frac {68}{158}\times 25$ grams

=10.76 grams

Hence, 25 grams of $KMnO_4$ has 10.76 grams of oxygen.