1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains 0.110 M HNO2 and 0.170 M NaNO2. How many moles of HNO2 and NaNO2 remain i

Question

1.00 mL of 12.0 M HCl is added to 1.00 L of a buffer that contains 0.110 M HNO2 and 0.170 M NaNO2. How many moles of HNO2 and NaNO2 remain in solution after addition of the HCl

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Hưng Khoa 3 years 2021-07-26T11:44:34+00:00 1 Answers 77 views 0

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    2021-07-26T11:46:16+00:00

    Answer:

    Moles of NaNO2 = 0.158

    Moles of HNO2 final = 0.098

    Explanation:

    Given

    Moles of HCl = 12

    Moles of HNO2 = 0.11

    Moles of NaNO2 = 0.170

    HCl +NaNO2 –> HNO2  + NaCl

    1 mole of HCl react with one mole of NaNO2 to produce 1 mole of NaCl and 1 mole of HNO2

    Moles of NaNO2 = 0.17 – 0.012 = 0.158

    Moles of HNO2 final = 0.11 – 0.012 = 0.098

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