Question

07 Question (a peoints) See page 122 0.300g of Fe3O4 reacts with excess O2 to give Fed, in535% yield according to the following balanced equation 4Fesod s) +02(g)-→ 6Fe2O3(s) 1st attempt See Periodic Table Part 1 (1point) Calculate the theoretical yield of Fe,O Part 2 (1point) Calculate the actual yield of Fe Os

Answer 1

1)

Molar mass of Fe3O4,

MM = 3*MM(Fe) + 4*MM(O)

= 3*55.85 + 4*16.0

= 231.55 g/mol

mass(Fe3O4)= 0.3 g

number of mol of Fe3O4,

n = mass of Fe3O4/molar mass of Fe3O4

=(0.3 g)/(231.55 g/mol)

= 1.296*10^-3 mol

Balanced chemical equation is:

4 Fe3O4 + O2 ---> 6 Fe2O3

Molar mass of Fe2O3,

MM = 2*MM(Fe) + 3*MM(O)

= 2*55.85 + 3*16.0

= 159.7 g/mol

According to balanced equation

mol of Fe2O3 formed = (6/4)* moles of Fe3O4

= (6/4)*0.001296

= 0.001943 mol

mass of Fe2O3 = number of mol * molar mass

= 1.943*10^-3*1.597*10^2

= 0.3104 g

Answer: 0.310 g

2)

% yield = actual mass*100/theoretical mass

53.6= actual mass*100/0.3104

actual mass=0.166 g

Answer: 0.166 g