HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Chemistry

02: Solutions

with Solutions - page 2
Qstn# B-6 Prvs-QstnNext-Qstn
  • #6
    How
    many mL of 0.1 M HCl are required to react completely with 1 g
    mixture of Na2CO3 and NaHCO3 containing equimolar amounts of both?
    Ans : Let the amount of
    Na2CO3 in the mixture be x g.
    Then, the amount of
    NaHCO3 in the mixture is (1 - x) g.
    Molar mass of Na2CO3 = 2 × 23 + 1 × 12 + 3 × 16
    = 106 g mol-1
    Number of moles Na2CO3
    Molar mass of NaHCO3 = 1 × 23 + 1 × 1 × 12 + 3 × 16
    = 84 g mol-1
    Number
    of moles of NaHCO3
    According to the
    question,

    ⇒ 84x =
    106 - 106x
    ⇒ 190x =
    106
    ⇒ x =
    0.5579
    Therefore, number of
    moles of Na2CO3
    = 0.0053 mol
    And, number of moles of
    NaHCO3
    = 0.0053 mol
    HCl reacts with Na2CO3 and NaHCO3 according to the following equation.


    1 mol of Na2CO3 reacts with 2 mol of HCl.
    Therefore, 0.0053 mol
    of Na2CO3 reacts with 2 × 0.0053 mol =
    0.0106 mol.
    Similarly, 1 mol of
    NaHCO3 reacts with 1 mol of HCl.
    Therefore, 0.0053 mol
    of NaHCO3 reacts with 0.0053 mol of HCl.
    Total moles of HCl
    required = (0.0106 + 0.0053) mol
    = 0.0159 mol
    In 0.1 M of HCl,
    0.1 mol of HCl is
    preset in 1000 mL of the solution.
    Therefore, 0.0159 mol
    of HCl is present in
    = 159 mL of the
    solution
    Hence, 159 mL of 0.1 M
    of HCl is required to react completely with 1 g
    mixture of Na2CO3 and NaHCO3, containing equimolar amounts of both.