Answered

IDNLearn.com makes it easy to find accurate answers to your questions. Our platform offers reliable and detailed answers, ensuring you have the information you need.

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

Sagot :

Sebassandinidn

Answer:

[tex]m_{KOH}= 142.0gKOH[/tex]

Explanation:

Hello there!

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

[tex]2MnO_2+4KOH+O_2\rightarrow 2KMnO4+2KOH+H_2[/tex]

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:

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

Best regards!

Your engagement is important to us. Keep sharing your knowledge and experiences. Let's create a learning environment that is both enjoyable and beneficial. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.