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10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.?
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10 g of a piece of marble was put into excess of dilute HCl . When the...
Problem:
Given, 10 g of marble and 1120 cm^3 of CO2 at STP.
To find:
The percentage of CaCO3 in the marble.
Solution:
The balanced chemical equation for the reaction between marble and HCl is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, we see that 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2. At STP, 1 mole of any gas occupies 22.4 L or 22400 cm^3.
Therefore, the number of moles of CO2 produced can be calculated as follows:
n = (1120 cm^3) / (22400 cm^3/mol) = 0.05 mol
From the equation, we see that 1 mole of CaCO3 produces 1 mole of CO2. Hence, the number of moles of CaCO3 in the marble can be calculated as follows:
n(CaCO3) = 0.05 mol
The molar mass of CaCO3 can be calculated as follows:
M(CaCO3) = M(Ca) + M(C) + 3M(O) = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol
Hence, the mass of CaCO3 in the marble can be calculated as follows:
m(CaCO3) = n(CaCO3) × M(CaCO3) = 0.05 mol × 100.09 g/mol = 5.005 g
Finally, the percentage of CaCO3 in the marble can be calculated as follows:
% CaCO3 = (m(CaCO3) / m(marble)) × 100% = (5.005 g / 10 g) × 100% = 50%
Therefore, the percentage of CaCO3 in the marble is 50%.
Given, 10 g of marble and 1120 cm^3 of CO2 at STP.
To find:
The percentage of CaCO3 in the marble.
Solution:
The balanced chemical equation for the reaction between marble and HCl is:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
From the equation, we see that 1 mole of CaCO3 reacts with 2 moles of HCl to produce 1 mole of CO2. At STP, 1 mole of any gas occupies 22.4 L or 22400 cm^3.
Therefore, the number of moles of CO2 produced can be calculated as follows:
n = (1120 cm^3) / (22400 cm^3/mol) = 0.05 mol
From the equation, we see that 1 mole of CaCO3 produces 1 mole of CO2. Hence, the number of moles of CaCO3 in the marble can be calculated as follows:
n(CaCO3) = 0.05 mol
The molar mass of CaCO3 can be calculated as follows:
M(CaCO3) = M(Ca) + M(C) + 3M(O) = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol
Hence, the mass of CaCO3 in the marble can be calculated as follows:
m(CaCO3) = n(CaCO3) × M(CaCO3) = 0.05 mol × 100.09 g/mol = 5.005 g
Finally, the percentage of CaCO3 in the marble can be calculated as follows:
% CaCO3 = (m(CaCO3) / m(marble)) × 100% = (5.005 g / 10 g) × 100% = 50%
Therefore, the percentage of CaCO3 in the marble is 50%.
Community Answer
10 g of a piece of marble was put into excess of dilute HCl . When the...
Are its simple ("better do it yourself")
when u will write the reaction
u will see
according to stoichiometry
" 1 mole of CaCO3 produces 1 mole of CO2"
so 0.1 mole of CaCO3 will produce 0.1 mole of CO2
but here 1120/22400=0.05 mole CO2 are only produced (which is only 50% of 0.1)
So percentage CaCO3 in the marble is 50%
when u will write the reaction
u will see
according to stoichiometry
" 1 mole of CaCO3 produces 1 mole of CO2"
so 0.1 mole of CaCO3 will produce 0.1 mole of CO2
but here 1120/22400=0.05 mole CO2 are only produced (which is only 50% of 0.1)
So percentage CaCO3 in the marble is 50%
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10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.?.
10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.?.
Solutions for 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? in English & in Hindi are available as part of our courses for NEET.
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10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.?, a detailed solution for 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? has been provided alongside types of 10 g of a piece of marble was put into excess of dilute HCl . When the reaction was completed,1120 cm^3 of CO2 was obtained at STP. What is the percentage of CaCo3 in the marble?? A)25% B)50% C)75% D)100% The correct answer is 50%.plz explain.? theory, EduRev gives you an
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