Question

6KI (aq) + 8HNO₂ (aq) →6KNO, (aq) + 2NO(g) +31₂(s) + 4H₂O(1) 2

a) Deduce which reagent is in excess.
b) Determine how many grams of this reactant will remain unreacted.
c) Determine how many grams of nitrogen monoxide, NO will be produced.​

Answer 1

a) From the balanced chemical equation, we can see that for every 6 moles of KI, only 8 moles of HNO2 are needed. Therefore, HNO2 is the reagent in excess.

b) We are given that HNO2 is in excess but not told by how much. To determine the amount that will remain unreacted, we need to know the amount of KI that reacted. If we know the moles of KI, we can calculate the moles of HNO2 required and subtract that from the total moles of HNO2 given. Then we can convert that to grams.

Assuming some amount of KI reacted (in moles), we can calculate:

* 6 moles KI reacted

* Requires 8 moles HNO2

* Total HNO2 given - 8 moles consumed = excess moles of HNO2

* Excess moles of HNO2 x molar mass of HNO2 = grams of HNO2 remaining unreacted

c) To determine the grams of NO produced, we can do:

* 2 moles NO produced for every 6 moles KI reacted

* Moles of KI reacted x 2/6 = moles of NO produced

* Moles of NO x molar mass of NO = grams of NO produced

So in summary, to solve the problem we need to know:

1) Moles of KI that reacted

2) Use that to calculate moles and grams of excess HNO2

3) Use moles of KI reacted to calculate moles and grams of NO produced

Hope this helps! Let me know if you have any other questions.