When nitric acid and potassium iodide react, they do so in accordance with 8 HNO

Question

When nitric acid and potassium iodide react, they do so in accordance with 8 HNO_3(aq) + 6 KI (aq) rightarrow 2 NO(g) + 3 I_2(s) + 4 H_2O(l) + 6 KNO_3(aq) when 450.0 mL of 0.375 M HNO_3(aq) and 200.0 mL of 0.440 M KI (aq) are mixed determine the following: Volume of the final solution in liters. The limiting reactant. The grams of I_2(s) that will be produced. The molar concentration of the un reacted aqueous iodide ions, [I-], in the final solution. The molar concentration (M) of aqueous nitrate ions, if the final mixture volume was reduced by 50%. The reducing and oxidizing agents.

Explanation / Answer

Moles of HNO3 = 0.45 L x 0.375 M == 0.16875 moles

Moles of KI = 0.2 L x 0.440 M == 0.088 moles

8 moles of HNO3 reacts with 6 moles of KI

So, 0.16875 moles reacts with (6/8) x 0.16875 moles = 0.1265 moles

But available moles is 0.088 moles .So, limiting reagent is KI

(c)

6 moles KI produce 3 moles I2

So, 0.088 moles KI produces (0.088 moles x (3/6)) = 0.044 moles

Molar mass of I2 = 253 .8 g/mol

So mass of I2 = 253.8 g/mol x 0.044 moles == 11.20 grams

Since KI is limiting reagent , all the I- ions will react and un reacted I- concentration is zero.

Here, oxidation number of Iodine changed from -1 to 0. It got oxidized .So, KI is reducing agent.

Here Nitrogen got reduced, so HNO3 is oxiding agent

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