Slesson 15 notes(3)( x
moles to liters conversio x
101 chem101
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question 10 of 20
how many liters of a 0.209 mki 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
culaq) + 12(s) +
4 kno3(aq)

volume in L = 0.25 L

Explanation:

Given data:

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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