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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 r...
Calculating moles of HNO3 and KOH
To determine the resulting concentration of KNO3 and KOH after the reaction between HNO3 and KOH, we first need to calculate the number of moles of each reactant. We can do this using the equation:
moles = concentration x volume
For HNO3:
moles HNO3 = 0.2 M x 0.08 L = 0.016 moles HNO3
For KOH:
moles KOH = 0.150 M x 0.12 L = 0.018 moles KOH
Determining the limiting reactant
To determine which reactant is the limiting reactant, we need to compare the number of moles of HNO3 and KOH. The reaction between HNO3 and KOH has a 1:1 mole ratio, which means that the limiting reactant will be the one with fewer moles. In this case, HNO3 has fewer moles (0.016 moles) than KOH (0.018 moles), so HNO3 is the limiting reactant.
Calculating the number of moles of KNO3 formed
Since HNO3 is the limiting reactant, all of it will be used up in the reaction. The balanced chemical equation shows that 1 mole of HNO3 reacts with 1 mole of KOH to produce 1 mole of KNO3 and 1 mole of H2O. Therefore, the number of moles of KNO3 formed will be equal to the number of moles of HNO3 used:
moles KNO3 = 0.016 moles HNO3
Calculating the concentration of KNO3
Now that we know the number of moles of KNO3 formed, we can calculate the concentration of KNO3 in the resulting solution using the equation:
concentration = moles / volume
The total volume of the resulting solution is 0.08 L + 0.12 L = 0.2 L. Therefore:
concentration KNO3 = 0.016 moles / 0.2 L = 0.0800 M KNO3
Calculating the concentration of KOH
Since KOH is not completely consumed in the reaction, we need to calculate its remaining concentration in the resulting solution. The balanced chemical equation shows that 1 mole of KOH reacts with 1 mole of HNO3. Therefore, the number of moles of KOH that react with HNO3 is also 0.016 moles. The remaining moles of KOH in the solution will be:
moles remaining KOH = 0.018 moles - 0.016 moles = 0.002 moles KOH
The concentration of KOH in the resulting solution will be:
concentration KOH = moles remaining KOH / volume
concentration KOH = 0.002 moles / 0.2 L = 0.0100 M KOH
Final answer
Therefore, the resulting solution after mixing 80.0 ml of 0.2 M HNO3 and 120.00 ml of 0.150
Community Answer
When 80.0 ml of 0.2 M HNO3 is added to 120.00 ml of 0.150 M KOH, the r...
80ml*0.2M=16miliMole HNO3 120mi*0.150M=18miliMole KOH . so 16 milimole of HNO3 reacts with 16milimole of kOH and 16 milimole of KNO3 will form and 2milimole of kOH will be excess. so if we convert milimole into M then the resulting solution is 0.016M KNO3 and 0.0200M KOH . so ,D is the answer.
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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 Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 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 Class 11 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?.
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 Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 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 Class 11 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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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?, a detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an
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