Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

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

Sagot :

sebassandin

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!

We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.