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2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)
    Correct answer is '7'. Can you explain this answer?
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    2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives ...
    Solution:

    Given, mass of FeSO4.xH2O = 2.78 g
    Mass of precipitate BaSO4 = 2.33 g
    We have to find the value of x.

    Step 1: Calculation of moles of BaSO4

    Molar mass of BaSO4 = 137 + 32 + 4(16) = 233 g/mol
    Moles of BaSO4 = mass/molar mass = 2.33/233 = 0.01 mol

    Step 2: Calculation of moles of FeSO4

    Molar mass of FeSO4.xH2O = 56 + 32 + (4+2)x16 = 56 + 32 + 6x16 = 56 + 32 + 96 = 184 + 8x
    Moles of FeSO4.xH2O = mass/molar mass = 2.78/(184 + 8x) mol

    Step 3: Calculation of moles of BaCl2

    Since BaCl2 is in excess, all the moles of BaSO4 formed are due to the reaction between FeSO4.xH2O and BaCl2.

    Number of moles of BaCl2 = number of moles of BaSO4 = 0.01 mol

    Step 4: Calculation of moles of FeSO4 reacting with BaCl2

    From the balanced chemical equation, we know that 1 mole of BaCl2 reacts with 1 mole of FeSO4 to form 1 mole of BaSO4.

    Therefore, moles of FeSO4 reacting with BaCl2 = 0.01 mol

    Step 5: Calculation of x

    From step 2, we know that moles of FeSO4.xH2O = 2.78/(184 + 8x) mol
    From step 4, we know that moles of FeSO4 reacting with BaCl2 = 0.01 mol

    Therefore, 2.78/(184 + 8x) = 0.01

    Solving for x, we get x = 7.

    Therefore, the value of x is 7.
    Free Test
    Community Answer
    2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives ...
    Molecular weight of hydrated ferrus sulphate= M; M= 152+18x.
    gm equi. of FeSO4.xH2O= BaSO4
    {  2.78/M}={2.33/233}�100
    M=278.
    152+18 x=278
    X=7.
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    2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)Correct answer is '7'. Can you explain this answer?
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    2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)Correct answer is '7'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about 2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)Correct answer is '7'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)Correct answer is '7'. Can you explain this answer?.
    Solutions for 2.78 g of FeSO4.xH2O on treatment with excess of BaCl2 solution gives 2.33 g of white precipitate of BaSO4. What is value of x? (Ba = 137, S = 32, Fe = 56, O = 16)Correct answer is '7'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
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