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When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the reaction HNO3 KOH ----> KNO3 H2O occurs and the resulting solution is (A) 0.0160 M KNO3 (B) 0.01800 M KNO3 (C) 0. 0800 M KNO3 and 0.0100 M KOH (D) 0.0160 M KNO3 and 0.0200 M KOH?
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Most Upvoted Answer
When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the r...
Given:

Volume of HNO3 solution = 80.0 mL

Concentration of HNO3 solution = 0.2 M

Volume of KOH solution = 120.00 mL

Concentration of KOH solution = 0.150 M

Chemical equation: HNO3 + KOH → KNO3 + H2O


Calculating moles of HNO3 and KOH:

Moles of HNO3 = Volume (in L) × Concentration = (80.0 mL ÷ 1000 mL/L) × 0.2 M = 0.016 mol

Moles of KOH = Volume (in L) × Concentration = (120.00 mL ÷ 1000 mL/L) × 0.150 M = 0.018 mol


Determining the limiting reactant:

To find the limiting reactant, we compare the moles of HNO3 and KOH. The reactant with the smaller number of moles is the limiting reactant.

In this case, HNO3 has 0.016 mol and KOH has 0.018 mol. Therefore, HNO3 is the limiting reactant.


Calculating moles of KNO3 produced:

From the balanced chemical equation, we can see that 1 mole of HNO3 reacts with 1 mole of KOH to produce 1 mole of KNO3.

Since HNO3 is the limiting reactant, the number of moles of KNO3 produced is equal to the number of moles of HNO3.

Therefore, moles of KNO3 = 0.016 mol


Calculating concentration of KNO3:

Concentration of KNO3 = Moles of KNO3 ÷ Total volume (in L) = 0.016 mol ÷ (80.0 mL + 120.00 mL) ÷ 1000 mL/L = 0.016 M


Calculating concentration of KOH:

Since KOH is not completely consumed in the reaction, there will be a remaining amount of KOH in the solution.

Volume of remaining KOH = Initial volume of KOH - Volume of KOH reacted = 120.00 mL - 80.0 mL = 40.00 mL

Concentration of remaining KOH = Concentration of KOH × Remaining volume (in L) = 0.150 M × (40.00 mL ÷ 1000 mL/L) = 0.006 M


Final Answer:

The resulting solution is 0.0160 M KNO3 and 0.0060 M KOH. Therefore, the correct option is (D) 0.0160 M KNO3 and 0.0200 M KOH.
Community Answer
When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the r...
Option c
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When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the reaction HNO3 KOH ----> KNO3 H2O occurs and the resulting solution is (A) 0.0160 M KNO3 (B) 0.01800 M KNO3 (C) 0. 0800 M KNO3 and 0.0100 M KOH (D) 0.0160 M KNO3 and 0.0200 M KOH?
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When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the reaction HNO3 KOH ----> KNO3 H2O occurs and the resulting solution is (A) 0.0160 M KNO3 (B) 0.01800 M KNO3 (C) 0. 0800 M KNO3 and 0.0100 M KOH (D) 0.0160 M KNO3 and 0.0200 M KOH? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the reaction HNO3 KOH ----> KNO3 H2O occurs and the resulting solution is (A) 0.0160 M KNO3 (B) 0.01800 M KNO3 (C) 0. 0800 M KNO3 and 0.0100 M KOH (D) 0.0160 M KNO3 and 0.0200 M KOH? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the reaction HNO3 KOH ----> KNO3 H2O occurs and the resulting solution is (A) 0.0160 M KNO3 (B) 0.01800 M KNO3 (C) 0. 0800 M KNO3 and 0.0100 M KOH (D) 0.0160 M KNO3 and 0.0200 M KOH?.
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