Question

4 A solution of 155 g of potassium iodide, KI is added to a solution of 175 g of nitric acid, $HNO_3$. The acid acts as an oxidizing agent. $6KI(aq) + 8HNO_3(aq) \rightarrow 6KNO_3(aq) + 2NO(g)$ $ + 3I_2(s) + 4H_2O(l)$ 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.

          4 A solution of 155 g of potassium iodide, KI is added to
a solution of 175 g of nitric acid, $HNO_3$. The acid acts
as an oxidizing agent.
$6KI(aq) + 8HNO_3(aq) \rightarrow 6KNO_3(aq) + 2NO(g)$
$ + 3I_2(s) + 4H_2O(l)$
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.

4 A solution of 155 g of potassium iodide, KI is added to
a solution of 175 g of nitric acid, HNO3. The acid acts
as an oxidizing agent.
6KI(aq) + 8HNO3(aq) → 6KNO3(aq) + 2NO(g)
+ 3I2(s) + 4H2O(l)
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.

Added by Kaylee P.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Vaidik Stats

Step 1

From the balanced equation, we can see that the mole ratio between KI and HNO3 is 6:8. This means that for every 6 moles of KI, we need 8 moles of HNO3. Show more…

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A solution of 15 g of nitric acid, HNO3. The acid acts as an oxidizing agent. 6KI(aq) + 8HNO3(aq) â†’ 6KNO3(aq) + 2NO(g) + 3I2(s) + 4H2O(l). a) Deduce which reagent is in excess. b) Determine how many grams of this reactant will remain unreacted. Monoxide, NO, will be produced.