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#1
What is the minimum mass of I2 that can be obtained from 50.0g KMnO4, 86.0g Kl and excess H2SO4?

Balanced Chemical reaction is:

2 KMnO4 + 10 Kl +8 H2SO4 ----> 6 K2SO4 + 2 MnSO4 + 5 I2 + 8H2O
0
2 years ago
#2
(Original post by Sugarcane44)
What is the minimum mass of I2 that can be obtained from 50.0g KMnO4, 86.0g Kl and excess H2SO4?

Balanced Chemical reaction is:

2 KMnO4 + 10 Kl ----> 6 K2SO4 + 2 MnSO4 + 5 I2 + 8H2O
I would suggest reviewing the equation above. There is no sulfur in the reactants and plenty amongst the products. The equation is balanced in terms of oxygen, hydrogen and potassium too.

When you have the balanced equation we can proceed
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#3
(Original post by TutorsChemistry)
I would suggest reviewing the equation above. There is no sulfur in the reactants and plenty amongst the products. The equation is balanced in terms of oxygen, hydrogen and potassium too.

When you have the balanced equation we can proceed
My bad, I have fixed the error
0
2 years ago
#4
(Original post by Sugarcane44)
My bad, I have fixed the error
Quick work!

I believe we must be looking for the maximum mass of I2 ? The minimum would be zero, as the reaction could be 0% efficient if we don't do it well.

The question already tells you that sulfuric acid is inxs. Hence you need to work out whether quantity of the KMnO4 or the quantity of KI is the limiting factor.
Work out how many moles you have of each. Obviously if you have more than 5 x more KI moles than KMnO4 then the permanganate is the limiting factor. If you have less than 5x more moles KI than KMnO4 then the iodide is the limiting factor.
Focus on the limiting compound for your calculation of the maximum number of moles of I2 that could be obtained. Then just calculate the mass of that number of moles of I2.
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