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22 April, 11:20

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Question 10 of 20

How many liters of a 0.209 M KI solution is needed to completely react with 2.43

g of Cu (NO3) 2 according to the balanced chemical reaction:

2

Cu (NO3) 2 (aq) + 4 Kl (aq) -

2

Cul (aq) + 1z (s) +

4 KNO3 (aq)

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Answers (1)
  1. 22 April, 11:41
    0
    volume in L = 0.25 L

    Explanation:

    Given dа ta:

    Mass of Cu (NO₃) ₂ = 2.43 g

    Volume of KI = ?

    Solution:

    Balanced chemical equation:

    2Cu (NO₃) ₂ + 4KI → 2CuI + I₂ + 4KNO₃

    Moles of Cu (NO₃) ₂:

    Number of moles = mass / molar mass

    Number of moles = 2.43 g / 187.56 g/mol

    Number of moles = 0.013 mol

    Now we will compare the moles of Cu (NO₃) ₂ with KI.

    Cu (NO₃) ₂ : KI

    2 : 4

    0.013 : 4 * 0.013=0.052 mol

    Volume of KI:

    Molarity = moles of solute / volume in L

    volume in L = moles of solute / Molarity

    volume in L = 0.052 mol / 0.209 mol/L

    volume in L = 0.25 L
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