Question

# TiO2(s) ?H

TiO2(s) ?H

TiO2 + 2C -> + Ti + 2CO Hf = -110.5

m = 12 kg of C

m = 45 kg TiO2

m = 7.5 kg Ti

a) limiting reactant and Theoretical Yield of Ti

for the limiting reactant, we need the balanced equation

TiO2 + 2C -> + Ti + 2CO

1 mol of TiO2 will react with 2 mol of C to form 1 mol of Ti and 2 mol of CO

change all masses to moles

m = 12 kg of C

n = m/MW = 12 kg of C / 12 kg/kmol of C = 1kmol of C

m = 45 kg TiO2

n = m/MW = 45 kg / 79.8 kg/kmol = 0.56 kmol of TiO2

Relationship is 1 mol of TiO2: 2 mol of CO

0.56 mol will need : 1.12 mol of C which we DONT have

Carbon is the limiting reactant!

Now calculating the theoretical yield

% yield = Actual Yield/Theoretical Yield * 100%

Theoretical Yield = if ALL the limiting reactant is going to be reacted to form the products, that is... if the 1 kmol of C is consumed... you should get:

calculate moles of Ti produced

n = m/MW = 7.5 kg of Ti / 47.86 kg/kmol = 0.156 kmol of Ti are produced

Lets get back to yield...

in theory... Ratio of 2 mol of C : 1 mol of Ti

But we have 1 kmol:0.5 kmol of Ti

The theoretical Yield should be 0.5 kmol of Ti (insted of the 0.15)

% yiel = real/theoretical *100% = 0.15/0.5 * 100% = 0.3*100 = 30%

% yield = 30%

b) Volume if P=1.1 bar and T = 50