NEET Exam > NEET Questions > 5 g of a piece of marble was treated with ex...
Start Learning for Free
5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]
- a)20%
- b)80%
- c)60%
- d)40%
Correct answer is option 'D'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
5 g of a piece of marble was treated with excess of dil. HCI. When re...
CaCO3 + 2HCI -> CaCI2 + 2H2O + CO2
Moles of CaCO3 = Moles of CO2
Mass of CaCO3 = 0.02 x 100 g = 2 g
∴ Percentage of CaCO3 
Free Test
FREE
| Start Free Test |
Community Answer
5 g of a piece of marble was treated with excess of dil. HCI. When re...
Given:
Mass of marble = 5 g
Volume of CO2 at STP = 448 ml
To find:
Percentage of CaCO3 in the marble
Step 1: Calculate the number of moles of CO2 produced.
Using the ideal gas equation, we can calculate the number of moles of CO2:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (448 ml = 0.448 L)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (STP = 273 K)
Plugging in the values:
(1 atm) (0.448 L) = n (0.0821 L.atm/mol.K) (273 K)
n = (1 atm * 0.448 L) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 0.020 mol
Step 2: Calculate the number of moles of CaCO3.
The balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
From the equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2.
Therefore, the number of moles of CaCO3 is also 0.020 mol.
Step 3: Calculate the molar mass of CaCO3.
Given the molar mass of CaCO3 is 100 g/mol.
Step 4: Calculate the mass of CaCO3.
Mass = number of moles * molar mass
Mass of CaCO3 = 0.020 mol * 100 g/mol
Mass of CaCO3 = 2 g
Step 5: Calculate the percentage of CaCO3.
Percentage = (mass of CaCO3 / mass of marble) * 100
Percentage = (2 g / 5 g) * 100
Percentage ≈ 40%
Therefore, the percentage of CaCO3 in the marble is approximately 40%.
Hence, the correct answer is option D.
Mass of marble = 5 g
Volume of CO2 at STP = 448 ml
To find:
Percentage of CaCO3 in the marble
Step 1: Calculate the number of moles of CO2 produced.
Using the ideal gas equation, we can calculate the number of moles of CO2:
PV = nRT
Where:
P = pressure (STP = 1 atm)
V = volume (448 ml = 0.448 L)
n = number of moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (STP = 273 K)
Plugging in the values:
(1 atm) (0.448 L) = n (0.0821 L.atm/mol.K) (273 K)
n = (1 atm * 0.448 L) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 0.020 mol
Step 2: Calculate the number of moles of CaCO3.
The balanced chemical equation for the reaction between CaCO3 and HCl is:
CaCO3 + 2HCl → CaCl2 + CO2 + H2O
From the equation, we can see that 1 mole of CaCO3 produces 1 mole of CO2.
Therefore, the number of moles of CaCO3 is also 0.020 mol.
Step 3: Calculate the molar mass of CaCO3.
Given the molar mass of CaCO3 is 100 g/mol.
Step 4: Calculate the mass of CaCO3.
Mass = number of moles * molar mass
Mass of CaCO3 = 0.020 mol * 100 g/mol
Mass of CaCO3 = 2 g
Step 5: Calculate the percentage of CaCO3.
Percentage = (mass of CaCO3 / mass of marble) * 100
Percentage = (2 g / 5 g) * 100
Percentage ≈ 40%
Therefore, the percentage of CaCO3 in the marble is approximately 40%.
Hence, the correct answer is option D.
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer?
Question Description
5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer?.
5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer?.
Solutions for 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer?, a detailed solution for 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? has been provided alongside types of 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 5 g of a piece of marble was treated with excess of dil. HCI. When reaction was complete 448 ml of CO2 was obtained at STP. The percentage of CaCO3 in the marble is [Molar mass of CaCO3 = 100 g/mol]a)20%b)80%c)60%d)40%Correct answer is option 'D'. Can you explain this answer? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup