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2 years ago

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

Chemistry

2 answers:

MArishka [77]2 years ago

4 0

Answer: 10.1 g

I just did it and got i right

Katen [24]2 years ago

3 0

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.

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What is the mole fraction of NaOH in an aqueous solution that

astraxan [27]

Answer:

The mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

Explanation:

We are given that

Aqueous solution that contains 22.9% NaOH by mass means

22.9 g NaOH in 100 g solution.

Mass of NaOH(WB)=22.9 g

Mass of water =100-22.9=77.1

Na=23

O=16

H=1.01

Molar mass of NaOH(MB)=23+16+1.01=40.01

Number of moles = $\frac{Given\;mass}{Molar\;mass}$

Using the formula

Number of moles of NaOH $(n_B)=\frac{W_B}{M_B}=\frac{22.9}{40.01}$

$n_B=0.572moles$

Molar mass of water=16+2(1.01)=18.02g

Number of moles of water $(n_A)=\frac{77.1}{18.02}$

$n_A=4.279 moles$

Now, mole fraction of NaOH

= $\frac{n_B}{n_B+n_A}$

$=\frac{4.279}{0.572+4.279}$

=0.882

Hence, the mole fraction of NaOH in an aqueous solution that contain 22.9% NaOH by mass=0.882

4 0

3 years ago

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Explanation:

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