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18 April, 04:09

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179 x 1023 atoms of magnesium with 54.21 g of phosphoric acid (H3PO4) ?

3 Mg + 2 H3 (PO4) - -> Mg3 (PO4) 2 + 3 H2

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  1. 18 April, 05:51
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    18.58 liters of hydrogen gas

    Explanation:

    We are given;

    The equation;

    3Mg + 2H₃ (PO₄) → Mg₃ (PO₄) ₂ + 3H₂

    Atoms of Magnesium = 7.179 x 10^23 atoms Mass of phosphoric acid as 54.21 g

    We are required to determine the volume of hydrogen gas produced;

    Step 1; moles of Magnesium

    1 mole of an element contains 6.02 * 10^23 atoms

    therefore;

    Moles of Mg = (7.179 x 10^23) : (6.02 * 10^23)

    = 1.193 moles

    Step 2: Moles of phosphoric acid

    moles = Mass : Molar mass

    Molar mass of phosphoric acid = 97.994 g/mol

    Therefore;

    Moles of Phosphoric acid = 54.21 g : 97.994 g/mol

    = 0.553 moles

    Step 3: Determine the rate limiting reagent

    From the mole ratio of Mg to Phosphoric acid (3 : 2);

    1.193 moles of magnesium requires 0.795 moles of phosphoric acid while,

    0.0553 moles of phosphoric acid requires 0.8295 moles of Mg

    Therefore, phosphoric acid is the rate limiting reagent

    step 4: Determine the moles of hydrogen produced

    From the equation, w moles of phosphoric acid reacts to produce 3 moles of hydrogen;

    Therefore; moles of Hydrogen = moles of phosphoric acid * 3/2

    = 0.553 moles * 3/2

    = 0.8295 moles

    Step 5: Volume of hydrogen gas

    1 mole of a gas occupies a volume of 22.4 liters at STP

    Therefore;

    Volume of Hydrogen = 0.8295 moles * 22.4 L/mol

    = 18.58 Liters

    Therefore; 18.58 liters of hydrogen gas will be produced
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