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
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