Question

Copper(II) sulfate reacts with sodium phosphate in a precipitation reaction. If 103.77 g of copper(II) sulfate is reacted with 172.33 g of sodium phosphate, what is the theoretical yield of copper(II) phosphate? Report your answer rounded to the hundredths of a gram without units.

Answer 1

Molar mass of CuSO4,

MM = 1*MM(Cu) + 1*MM(S) + 4*MM(O)

= 1*63.55 + 1*32.07 + 4*16.0

= 159.62 g/mol

mass(CuSO4)= 103.77 g

use:

number of mol of CuSO4,

n = mass of CuSO4/molar mass of CuSO4

=(1.038*10^2 g)/(1.596*10^2 g/mol)

= 0.6501 mol

Molar mass of Na3PO4,

MM = 3*MM(Na) + 1*MM(P) + 4*MM(O)

= 3*22.99 + 1*30.97 + 4*16.0

= 163.94 g/mol

mass(Na3PO4)= 172.33 g

use:

number of mol of Na3PO4,

n = mass of Na3PO4/molar mass of Na3PO4

=(1.723*10^2 g)/(1.639*10^2 g/mol)

= 1.051 mol

Balanced chemical equation is:

3 CuSO4 + 2 Na3PO4 ---> Cu3(PO4)2 + 3 Na2SO4

3 mol of CuSO4 reacts with 2 mol of Na3PO4

for 0.6501 mol of CuSO4, 0.4334 mol of Na3PO4 is required

But we have 1.051 mol of Na3PO4

so, CuSO4 is limiting reagent

we will use CuSO4 in further calculation

Molar mass of Cu3(PO4)2,

MM = 3*MM(Cu) + 2*MM(P) + 8*MM(O)

= 3*63.55 + 2*30.97 + 8*16.0

= 380.59 g/mol

According to balanced equation

mol of Cu3(PO4)2 formed = (1/3)* moles of CuSO4

= (1/3)*0.6501

= 0.2167 mol

use:

mass of Cu3(PO4)2 = number of mol * molar mass

= 0.2167*3.806*10^2

= 82.47 g

Answer: 82.5 g