Question

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

Answer 1

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:

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

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