Before solving the problem write a balanced
chemical equation
Excess aqueous NaI was added to 50 mL
of aqueous CuNO3 solution, and 14.2 g of
CuI precipitate was formed. What was the
molar concentration of CuNO3 in the original
solution?
Answer in units of M
To solve this problem, we need to write a balanced chemical equation to understand the stoichiometry involved in the reaction between NaI and CuNO3. The reaction can be represented as follows:
2NaI + Cu(NO3)2 -> 2NaNO3 + CuI
From the balanced equation, we can see that 2 moles of NaI react with 1 mole of Cu(NO3)2 to form 2 moles of NaNO3 and 1 mole of CuI.
Now let's calculate the number of moles of CuI formed using the given mass of CuI precipitate. We can use the molar mass of CuI to convert grams of CuI to moles:
Molar mass of CuI = atomic mass of Cu + atomic mass of I
= (63.55 g/mol) + (126.90 g/mol)
= 190.45 g/mol
Moles of CuI = Mass of CuI / Molar mass of CuI
= 14.2 g / 190.45 g/mol
= 0.0748 mol
Since 1 mole of Cu(NO3)2 produces 1 mole of CuI, the number of moles of Cu(NO3)2 in the original solution is also 0.0748 mol.
Now let's find the molar concentration of CuNO3 in the original solution. We'll need the volume of the original solution, which is 50 mL. However, the concentration must be in units of M, which is mol/L. Therefore, we need to convert mL to L:
Volume of original solution = 50 mL = 0.050 L
Molar concentration (Molarity) of CuNO3 = Moles of CuNO3 / Volume of solution in liters
= 0.0748 mol / 0.050 L
= 1.496 M
Therefore, the molar concentration of CuNO3 in the original solution is approximately 1.496 M.