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4 April, 15:53

How many milliliters of 3.00 M H2SO4 are required to react with 4.35 g of solid containing 23.2 wt% Ba (NO3) 2 if the reaction is Ba21 1 SO22 n BaSO (s) ?

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  1. 4 April, 19:35
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    1.29 mL.

    Explanation:

    Equation of the reaction:

    H2SO4 (aq) + Ba (NO3) 2 (aq) - -> BaSO4 + 2HNO3 (aq)

    Mass of Ba (NO3) 3 = wt% * mass of the solid

    = 23.2 x 4.35 / 100

    = 1.01 g

    Number of moles of Ba (NO3) 2 = mass/molar mass

    Molar mass of Ba (NO3) 2 = 137 + (14 + (16 * 3)) * 2

    = 261 g/mol

    = 1.01 / 261

    = 0.00387

    By stoichiometry, since 1 mole of Ba (NO3) 2 reacted with 1 mole of H2SO4. Therefore, number of moles of H2SO4 = 0.00387 moles.

    Volume = number of moles/molar concentration

    = 0.00387/3

    = 0.00129 L

    = 1.29 mL.
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