All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >  Reaction of x g of Sn with HCl quantitatively... Start Learning for Free
SaveJoin the Discussion on the App
Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).
(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)
Q. The value of y is _______.
    Correct answer is '1.23'. Can you explain this answer?
    FREE This question is part of
    Download PDF Attempt Test
    		Download PDF Attempt this Test
    Most Upvoted Answer
    Reaction of x g of Sn with HCl quantitatively produced a salt. Entire ...
    Given information:
    - x g of Sn reacted quantitatively with HCl to produce a salt
    - The entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively)
    - Molar masses (in g mol-1) of H, C, N, O, Cl and Sn are 1, 12, 14, 16, 35 and 119, respectively

    To find: Value of y

    Solution:
    1. Finding the moles of Sn in x g of Sn:
    - Molar mass of Sn = 119 g/mol
    - Moles of Sn = mass/molar mass = x/119
    - Let's assume this value as 'a'

    2. Finding the moles of HCl required to react with 'a' moles of Sn:
    - From the balanced chemical equation of the reaction between Sn and HCl, we know that 1 mole of Sn reacts with 2 moles of HCl
    - So, 'a' moles of Sn will react with 2*a moles of HCl
    - Molar mass of HCl = 35 + 1 = 36 g/mol
    - Mass of HCl required = moles * molar mass = 2*a*36 = 72a
    - Let's assume this value as 'b'

    3. Finding the moles of the salt produced in step 2:
    - Since the reaction between Sn and HCl was quantitative, all the Sn reacted to form the salt
    - So, the moles of the salt produced will be equal to the moles of Sn used, which is 'a'

    4. Finding the moles of nitrobenzene required to react with 'a' moles of the salt:
    - From the given information, we know that the entire amount of the salt reacted with y g of nitrobenzene
    - Molar mass of nitrobenzene = 12*7 + 1*5 + 14 = 93 g/mol
    - Moles of nitrobenzene required = mass/molar mass = y/93
    - Let's assume this value as 'c'

    5. Finding the moles of the organic salt produced in step 4:
    - From the given information, we know that 1.29 g of the organic salt was produced
    - Molar mass of the organic salt is not given, so let's assume it as 'M'
    - Moles of the organic salt produced = mass/molar mass = 1.29/M
    - Let's assume this value as 'd'

    6. Writing the balanced chemical equation of the reaction between the salt and nitrobenzene:
    - Let's assume that the salt is SnCl2.X, where X represents the anion part of the salt
    - In the presence of HCl, the salt reacts with nitrobenzene to form an organic salt, which can be represented as:
    SnCl2.X + C6H5NO2 + HCl -> Organic salt + SnCl4 + H2O
    - Since the reaction was quantitative, all the salt reacted with nitrobenzene to form the organic salt

    7. Finding the moles of SnCl2 in the salt:
    - Let's assume that the molar mass
    Free Test
    FREE
    Start Free Test
    		Start Free Test
    Community Answer
    Reaction of x g of Sn with HCl quantitatively produced a salt. Entire ...
    1 mole of organic salt is produced by 1 mole of nitrobenzene 0.01 mole of organic salt is produced by 0.01 mole nitrobenzene.
    Molar mass of nitrobenzene
    = 12 × 6 + 1 × 5 + 14 × 1 + 16 × 2
    = 72 + 5 + 14 + 32 = 123 g mol-1
    Mass of nitrobenzene required,
    y = moles of nitrobenzene × molar mass
    = 0.01 × 123 = 1.23 g
    < Previous Question Next Question >
    Explore Courses for JEE exam

    Similar JEE Doubts

    Consider a solution of CH3COONH4 which is a salt of weak acid weak base. Theequilibrium involved in the solutions are :If we add these three reactions, then the net reaction isBoth CH3C00- and NH4 get hydrolysed independently and their hydrolysis depends on(i) their initial concentration(ii) the value of Kh which is Since both of the ions were produced from the same salt, their initial concentrations are same . Therefore unless untial the value of Kw/Kaand Kb is same, the degree of hydrolysis of ion cant be same.To explain why we assume that degree of hydrolysis of cation and anion is same, we need to now look at the third reaction i.e., combination of H+ and OH ions. It is obvious that this reaction happens only because one reaction produced H+ ion and the other produced OH ions. We can also note that this reaction causes both the hydrolysis reaction to occur more since their product ions are being consumed. Keep this thing in mind that the equilibrium which has smaller value of equilibrium conxtant is affected more by the common ion effect. For the same reason if for any reason a reaction is made to occur to a greater extent by the consumption of any one of the product ion, the reaction with the smaller value of equilibrium constant tends to get affected more.Therefore we conclude that firstly the hydrolysis of both the ions ocurs more in the presence of each other (due to consumption of the product ions) than in each other is absence. Secondly the hydrolysis of the ion which occurs to a lesser extent (due to smaller value of Kh) is affected more than the one whose Kh is greater. Hence we can see that the degree of hydrolysis of both the ions would be close to each other when they are getting hydrolysed in the presence of each other.Q.In a solution of NaHCO3 , the amphiprotic anion can under ionization to form H+ ion and hydrolysis to from OH ion.To calculat PH, suitable approximation is

    The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumq.Above homogeneous reaction is carried out in a 2 litre container at a particular temperature by taking 1 mole each of A, B, C and D respectively. If Kc for the reaction is then equilibrium concentration of C is

    The rate of chemical reaction at a particular temperature is proportional to the product of the molar concentration of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactant taking part in the reaction. aA + bB products, rate of reaction a [Aka [Bib = k [Aka [Bib, where k is the rate constant of the reaction.EQUILIBRIUM CONSTANT (k) For a general reaction aA + bB = cC + dD, forward rate rf = kf [A]a [B]b, backward rate rb = kb [Cicpyi ,concentrations of reactants products at equilibrium are related bywhere all theconcentrations are expressed in mole/liter.In the expression of equilibrium constant those components are kept whose concentration changes with time.If equilibrium is estabilished by taking all the components in the reaction then to predict the direction of reactions we calculate the reaction quotient (Q).The values of expression at any time during reaction is called reaction quotient if Q Kc reaction proceed in backward direction until equilibrium in reached if Q Kc reaction will proceed in forward direction until equilibrium is established if Q = KC Reaction is at equilibriumQ.When C,H5OH and CH3COOH are mixed in equivalent proportion equilibrium is reached when 2/3 of acid and alcohol are used. How much ester will be present when 2 moles of acid were to react with 2 moles of alcohol? .

    Top Courses for JEEView all

    Top Courses for JEE

    Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer?
    Question Description
    Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer?.
    Solutions for Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
    Here you can find the meaning of Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer?, a detailed solution for Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? has been provided alongside types of Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Reaction of x g of Sn with HCl quantitatively produced a salt. Entire amount of the salt reacted with y g of nitrobenzene in the presence of required amount of HCl to produce 1.29 g of an organic salt (quantitatively).(Use Molar masses (in g mol-1) of H, C, N, O, Cl and Sn as 1, 12, 14, 16, 35 and 119, respectively)Q. The value of y is _______.Correct answer is '1.23'. Can you explain this answer? tests, examples and also practice JEE tests.
    Explore Courses for JEE exam

    Top Courses for JEE

    Explore Courses

    Suggested Free Tests

    Signup for Free!
    Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
    10M+ students study on EduRev
    Get App
    © EduRev
    Education Revolution
    Signup to see your scores go up within 7 days!
    Access 1000+ FREE Docs, Videos and Tests
    Takes less than 10 seconds to signup