Question

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

Answer 1

`m_(KOH)= 142.0gKOH`

Step-by-step 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*(1molKMnO_4)/(158.03gKMnO_4)*(4molKOH)/(2molKMnO_4) *(56.11gKOH)/(1molKOH)\\\\m_(KOH)= 142.0gKOH`

Best regards!