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1 July, 03:59

Elemental phosphorus reacts with chlorine gas according to the equation: P4 (s) + 6Cl2 (g) ❝4PCl3 (l) A reaction mixture initially contains 45.55g P4 and 131.9g Cl2.

Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant is left?

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  1. 1 July, 04:05
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    So after reaction completion. 7.11 g of P₄ will be left over.

    Explanation:

    So for number of moles of P₄ = Given mass / molar mass = 45.55 / 123.88

    = 0.367 moles

    Number of moles of Cl₂ = Given mass / molar mass = 131.9/70 = 1.884 moles

    We will divide the values by dividing excess of reactants with their coefficient for finding excess of reactants.

    P₄ = 0.367 / coefficient 1 = 0.367

    Cl₂ = 1.884 / 6 = 0.314

    Phosphorus is thus the excess reactant.

    We need to use the mole value of the limiting reactant in order to find the mass of P₄ used. Using dimensional analysis,

    1.884 mol Cl₂ (1mole P₄/6 mol Cl₂) (123.88 g P₄ / 1 mol P₄) = 38.44 g P₄

    This is the amount which is used up. So by subtracting this amount from the Initial amount

    45.55 - 38.44 = 7.11 g of P₄

    So after reaction completion. 7.11 g of P₄ will be left over.
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