NEET Exam > NEET Questions > Excess of 2moles/dm3 HCL is treated with 2.8g...
Start Learning for Free
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is?
Most Upvoted Answer
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex F...
Calculation of Volume of HCL Consumed
To calculate the volume of HCL consumed, we need to follow the stoichiometry of the given reaction:
Fe + 2HCl → FeCl2 + H2
From the balanced chemical equation, we can see that 1 mole of iron (Fe) reacts with 2 moles of hydrochloric acid (HCl) to produce 1 mole of iron(II) chloride (FeCl2) and 1 mole of hydrogen gas (H2).
Given that 2 moles/dm3 of HCl is used, we need to determine the number of moles of Fe used in the reaction. We can use the following formula:
Number of moles = mass / molar mass
The molar mass of Fe is 55.845 g/mol. Therefore, the number of moles of Fe used is:
Number of moles of Fe = 2.8 g / 55.845 g/mol = 0.0501 moles
Since the stoichiometry of the reaction is 1:2 for Fe and HCl, the number of moles of HCl used is twice the number of moles of Fe used:
Number of moles of HCl = 2 x 0.0501 moles = 0.1002 moles
Now, we can use the following formula to calculate the volume of HCl consumed:
Volume of HCl consumed (in dm3) = number of moles / molarity
Since the molarity of HCl is 2 moles/dm3, the volume of HCl consumed is:
Volume of HCl consumed = 0.1002 moles / 2 moles/dm3 = 0.0501 dm3 or 50.1 mL
Explanation
The given reaction is a single displacement reaction, where iron reacts with hydrochloric acid to produce iron(II) chloride and hydrogen gas. The balanced chemical equation shows that 1 mole of iron reacts with 2 moles of hydrochloric acid. Therefore, to determine the number of moles of HCl consumed, we need to know the number of moles of Fe used in the reaction. This can be calculated from the given mass of Fe and its molar mass. Once we know the number of moles of Fe used, we can use the stoichiometry of the reaction to determine the number of moles of HCl consumed. Finally, the volume of HCl consumed can be calculated using its molarity and the number of moles consumed.
Community Answer
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex F...
I got it. for 0.05 moles of fecl2 there must be 0.1 mole of hcl(by stoichiometry) which on equating[density(mole/volume)2=0.1/volume] with 2 mole/dm3 gives 0.05 dm3 volume that is 50 ml.
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
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is?
Question Description
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is?.
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is?.
Solutions for Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? 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 Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? defined & explained in the simplest way possible. Besides giving the explanation of
Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is?, a detailed solution for Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? has been provided alongside types of Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? theory, EduRev gives you an
ample number of questions to practice Excess of 2moles/dm3 HCL is treated with 2.8gm of Fe as shown in rex Fe+HCL --FeCl2+H2 The volume of HCL consumed is? 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