Problem: What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO 3)2? 2 Cu(NO3)2 + 4 KI ⟶ 2 CuI + I 2 + 4 KNO3

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FREE Expert Solution

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


Recall Molarity:

Molarity (M)=moles soluteL solution


Step 1: Calculate moles of Cu(NO3)2 required to react with KI.

moles Cu(NO3)2=0.3842mol Cu(NO3)2L×(43.88 mL×103 L1 mL)

moles Cu(NO3)2 = 0.01686 mol


Step 2: Determine the moles of KI using mole-to-mole comparison (from the net balanced reaction).

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Problem Details

What volume of a 0.2089 M KI solution contains enough KI to react exactly with the Cu(NO3)2 in 43.88 mL of a 0.3842 M solution of Cu(NO 3)2

2 Cu(NO3)2 + 4 KI ⟶ 2 CuI + I 2 + 4 KNO3

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Based on our data, we think this problem is relevant for Professor Giles' class at MC MARICOPA.

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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.