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2 April, 08:37

Calculate the volume of the chlorine that would be required to react completely with 3.70g of dry slaked lime according to the equation Ca (OH) 2+Cl=CaOCl2+H2O

H = 1, O=16, CA=40,1 mole of gas occupies 22.4dm3 at s. t. p

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  1. 2 April, 09:00
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    1.12 dm³

    Explanation:

    Given dа ta:

    Volume of chlorine = ?

    Mass of slaked lime = 3.70 g

    Solution:

    Chemical equation;

    Ca (OH) ₂ + Cl₂ → CaOCl₂ + H₂O

    Number of moles of slaked lime:

    Number of moles = mass / molar mass

    Number of moles = 3.70 g / 74.1 g/mol

    Number of moles = 0.05 mol

    Now we will compare the moles of Ca (OH) ₂ and chlorine.

    Ca (OH) ₂ : Cl₂

    1 : 1

    0.05; 0.05

    Volume of chlorine:

    volume of one mole of gas at stp = 22.4 dm³

    0.05 * 22.4 dm³ = 1.12 dm³
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