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

The correct answer is - 0.570 grams

Step-by-step 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