Question

Write a balanced reaction for each reaction. 0.4M NaI + 0.1M KMnO4 + 3M H2SO4 0.4M NaI + 0.1M FeCl3 + 3M H2SO4 0.4M NaI...

Write a balanced reaction for each reaction.

0.4M NaI + 0.1M KMnO4 + 3M H2SO4

0.4M NaI + 0.1M FeCl3 + 3M H2SO4

0.4M NaI + 0.1M KMnO4 + 6M NaOH

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Answer #1

1) The general equation for the reaction is :

4KMnO4 + 10NaI + 16H2SO4 <---> 5I2 + 4MnSO4 + 10NaSO4 + 2K2SO4 + 16H2O

In the given question, we have 0.4 moles NaI, 0.1 moles KMnO4 and 3 moles H2SO4.

Here KMnO4 is the limiting reagent, hence the equation becomes

0.4 NaI + 0.1 KMnO4 + 3 H2SO4 <---> 0.125 I2 + 0.1 MnSO4 + 0.25 NaSO4 + 0.05 K2SO4 + 0.4 H2O

2) The general reaction is:

2NaI + 2 FeCl3 + 3 H2SO4 <----> I2 + 2 FeSO4 + Na2SO4 + 6 HCl

According to given equation, the limiting reagent is FeCl3

So the solution is:

0.4 NaI + 0.1 FeCl3 + 3 H2SO4 <----> 0.05 I2 + 0.1 FeSO4 + 0.05 Na2SO4 + 0.3 HCl

3) The general reaction is:

20 NaI + 80 KMnO4 + 60 NaOH <---> 20 IO3 + 3 H​​​​​​20 + 40 K2MnO4 + 40 Na2MnO4

According to given question, the limiting reagent is NaOH

So the solution is:

0.4 NaI + 0.1 KMnO4 + 6 NaOH <---> 0.033 IO3 + 0.005 H​​​​​​20 + 0.0667 K2MnO4 + 0.0667 Na2MnO4

To find the limiting regent, we take one stoichiometric coefficient of one reactant and find the coefficients of others using the stoichiometry and find out which of the reactants are provided in lower quantity. It is the limiting reagent. If none of the other reactants is present in lower quantity, this means that the selected compound is the limiting reagent.

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Write a balanced reaction for each reaction. 0.4M NaI + 0.1M KMnO4 + 3M H2SO4 0.4M NaI + 0.1M FeCl3 + 3M H2SO4 0.4M NaI...
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