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rjkz [21]
3 years ago
14

If your end product is 200.0 g KMnO4 how much KOH did you start with?

Chemistry
1 answer:
aniked [119]3 years ago
8 0

Answer:

m_{KOH}= 142.0gKOH

Explanation:

Hello there!

In this case, according to the following chemical reaction we found on goo gle as it was not given:

2MnO_2+4KOH+O_2\rightarrow 2KMnO4+2KOH+H_2

Whereas we can see a 2:4 mole ratio of potassium permanganate product to potassium hydroxide reactant with molar masses of 158.03 g/mol and 54.11 g/mol respectively. In such a way, by developing the following stoichiometric setup, we obtain the mass of KOH to start with:

m_{KOH}=200.0gKMnO_4*\frac{1molKMnO_4}{158.03gKMnO_4}*\frac{4molKOH}{2molKMnO_4}  *\frac{56.11gKOH}{1molKOH}\\\\m_{KOH}= 142.0gKOH

Best regards!

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