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19 October, 10:10

How many mL of 0.563 M HNO3 are needed to dissolve 7.83 g of BaCoz? 2HNO3 (aq) + BaCO3 (s) - Ba (NO3) 2 (aq) + H2O (1) + CO2 (8) mL

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  1. 19 October, 10:21
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    Volume of HNO₃ required = 140 mL

    Explanation:

    Given dа ta:

    Molarity of HNO₃ = 0.563 M

    mass of BaCO₃ = 7.83 g

    Volume of HNO₃ = ?

    Solution:

    First of all we will write the balance chemical equation

    2HNO₃ + BaCO₃ → Ba (NO₃) ₂ + H₂O + CO₂

    Number of moles of BaCO₃ = mass / molar mass

    Number of moles of BaCO₃ = 7.83 g / 197.34 g/mol

    Number of moles of BaCO₃ = 0.04 mol

    Now we compare the moles of BaCO₃ and HNO₃.

    BaCO₃ : HNO₃

    1 : 2

    0.04 : 2*0.04 = 0.08 mol

    Volume of HNO₃ required = number of moles / Molarity

    Volume of HNO₃ required = 0.08 mol / 0.563 mol/L

    Volume of HNO₃ required = 0.14 L

    0.14 * 1000 = 140 mL
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