Question

One step in the isolation of pure rhodium metal (Rh) is the precipitation of rhodium(III) hydroxide from a solution containing rhodium(III) sulfate according to the following balanced chemical equation:

Rh2(SO4)3(aq) + 6NaOH(aq)  →  2Rh(OH)3(s) + 3Na2SO4(aq)

What is the theoretical yield of rhodium(III) hydroxide from the reaction of 0.540 g of rhodium(III) sulfate with 0.209 g of sodium hydroxide?

Answer 1

Molar mass of Rh2(SO4)3 = 2*MM(Rh) + 3*MM(S) + 12*MM(O)

= 2*102.9 + 3*32.07 + 12*16.0

= 494.01 g/mol

mass of Rh2(SO4)3 = 0.54 g

we have below equation to be used:

number of mol of Rh2(SO4)3,

n = mass of Rh2(SO4)3/molar mass of Rh2(SO4)3

=(0.54 g)/(494.01 g/mol)

= 1.093*10^-3 mol

Molar mass of NaOH = 1*MM(Na) + 1*MM(O) + 1*MM(H)

= 1*22.99 + 1*16.0 + 1*1.008

= 39.998 g/mol

mass of NaOH = 0.209 g

we have below equation to be used:

number of mol of NaOH,

n = mass of NaOH/molar mass of NaOH

=(0.209 g)/(39.998 g/mol)

= 5.225*10^-3 mol

we have the Balanced chemical equation as:

Rh2(SO4)3 + 6 NaOH ---> 2 Rh(OH)3 + 3 Na2SO4

1 mol of Rh2(SO4)3 reacts with 6 mol of NaOH

for 1.093*10^-3 mol of Rh2(SO4)3, 6.559*10^-3 mol of NaOH is required

But we have 5.225*10^-3 mol of NaOH

so, NaOH is limiting reagent

we will use NaOH in further calculation

Molar mass of Rh(OH)3 = 1*MM(Rh) + 3*MM(O) + 3*MM(H)

= 1*102.9 + 3*16.0 + 3*1.008

= 153.924 g/mol

From balanced chemical reaction, we see that

when 6 mol of NaOH reacts, 2 mol of Rh(OH)3 is formed

mol of Rh(OH)3 formed = (2/6)* moles of NaOH

= (2/6)*5.225*10^-3

= 1.742*10^-3 mol

we have below equation to be used:

mass of Rh(OH)3 = number of mol * molar mass

= 1.742*10^-3*1.539*10^2

= 0.268 g

Answer: 0.268 g