Question

If the NaOH is added to 35.0 mL of 0.167 M Cu(NO3)2 and the precipitate isolated by filtration, what is the theoretical yield of the reaction?

Answer 1

Answer:

The correct answer is - 0.570 grams

Explanation:

The balanced chemical reaction is given by

Cu(NO3)2(aq)     + 2NaOH(aq)    -------->    Cu(OH)2(s)      + 2NaNO3(aq)

        1.0 mole            2.0 mole                 1.0 mole          2.0 mole

number of mol of Cu(OH)2,

n = Molarity * Volume

= $35.0*0.167 = 5.845$ millimoles

As clear in the equation, 1 mole of Cu(NO3)2 gives 1 mole of Cu(OH)2 , So, 5.845 millimoles of Cu(NO3)2 will produce 5.845 millimoles of Cu(OH)2

Mass of Cu(OH)2 = number of mol * molar mass

= $97.5*5.845*10^-3$

= 0.570 grams

Thus, the correct answer is - 0.570 grams