You mix 2.0 mL of 0.50 M Cu(NO3)2 with 5.0 mL of 0.50 M KI, and collect and dry the CuI precipitate. Calculate the following:
a) Moles of Cu(NO3)2 used
b) Moles of KI used
c) Moles of CuI precipitate expected from Cu(NO3)2
d) Moles of CuI precipitate expected from KI
To solve this problem, you will need to use the concept of stoichiometry. Stoichiometry allows you to determine the relationship between the amount of reactants and products in a chemical reaction.
First, let's write the balanced chemical equation for the reaction:
Cu(NO3)2 + 2KI -> 2KNO3 + CuI
a) Moles of Cu(NO3)2 used:
We are given the volume and molarity of Cu(NO3)2. To calculate the moles, we can use the following formula:
moles = volume (in liters) x concentration (in M)
Given:
Volume = 2.0 mL = 0.002 L
Concentration = 0.50 M
moles = 0.002 L x 0.50 M
moles = 0.001 mol
Therefore, 0.001 moles of Cu(NO3)2 are used.
b) Moles of KI used:
Similarly, we can calculate the moles of KI using the given volume and molarity.
Given:
Volume = 5.0 mL = 0.005 L
Concentration = 0.50 M
moles = 0.005 L x 0.50 M
moles = 0.0025 mol
Therefore, 0.0025 moles of KI are used.
c) Moles of CuI precipitate expected from Cu(NO3)2:
From the balanced chemical equation, we see that 1 mole of Cu(NO3)2 reacts to form 1 mole of CuI. Therefore, the moles of CuI precipitate formed will be equal to the moles of Cu(NO3)2 used.
moles of CuI precipitate = moles of Cu(NO3)2 = 0.001 mol
Therefore, 0.001 moles of CuI precipitate are expected from Cu(NO3)2.
d) Moles of CuI precipitate expected from KI:
From the balanced chemical equation, we see that 2 moles of KI react to form 1 mole of CuI. Therefore, the moles of CuI precipitate formed will be half of the moles of KI used.
moles of CuI precipitate = (moles of KI) / 2 = 0.0025 mol / 2
moles of CuI precipitate = 0.00125 mol
Therefore, 0.00125 moles of CuI precipitate are expected from KI.
To calculate the moles of Cu(NO3)2 used, we can use the formula:
moles = concentration (M) × volume (L)
a) Moles of Cu(NO3)2 used:
Given:
Concentration of Cu(NO3)2 = 0.50 M
Volume of Cu(NO3)2 = 2.0 mL = 2.0/1000 L
Using the formula:
moles = 0.50 M × 2.0/1000 L = 0.001 mol
Therefore, 0.001 moles of Cu(NO3)2 were used.
b) Moles of KI used:
Given:
Concentration of KI = 0.50 M
Volume of KI = 5.0 mL = 5.0/1000 L
Using the formula:
moles = 0.50 M × 5.0/1000 L = 0.0025 mol
Therefore, 0.0025 moles of KI were used.
c) Moles of CuI precipitate expected from Cu(NO3)2:
According to the balanced chemical equation:
Cu(NO3)2 + 2KI -> CuI + 2KNO3
The stoichiometry of the reaction shows that 1 mole of Cu(NO3)2 reacts with 1 mole of CuI.
Since 0.001 moles of Cu(NO3)2 were used, we can expect the same amount of moles of CuI to form.
Therefore, 0.001 moles of CuI precipitate are expected from Cu(NO3)2.
d) Moles of CuI precipitate expected from KI:
According to the balanced chemical equation:
Cu(NO3)2 + 2KI -> CuI + 2KNO3
The stoichiometry of the reaction shows that 2 moles of KI react with 1 mole of CuI.
Since 0.0025 moles of KI were used, we can calculate the moles of CuI precipitate using the stoichiometry:
moles of CuI = 0.0025 moles of KI × (1 mole of CuI/2 moles of KI)
moles of CuI = 0.00125 mol
Therefore, 0.00125 moles of CuI precipitate are expected from KI.