Question

Jump to numerical question 48.34 kg C and 179.2 kg TiO2 react according to the equation: TiO2(s) +2C(s) Ti(s) + 2CO. 62.40 kg of Ti(s) is actually produced. What is the percent yield? (Do not include % sign in answer.) Submit response

Answer 1

TiO2 +2C---> Ti + 2CO
given
mass of C = 48.34 kg = 48340 g
Maas of TiO2 =179.2 kg = 179200 g

molar mass of C= 12.01 g/mol
So number of mols of C present = Mass of C/molar mass = 48340 g/12.01 g/mol
= 4024.98 mols

molar mass of TiO2 = 1*47.87 g/mol + 2*16 g/mol =79.87 g/mol
So number of mole of TiO2 present = 179200 g/79.87 g/mol
= 2,243.646 mols
SInce mole ratio of TiO2:C = 1:2

2,243.646 mols TiO2 need [2* 2,243.646 mols] C
=4487.292 mols of C.
But only 4024.98 mols C is present .Hence C will bethe limiting reactant.The reaction will
continue until all C is consumed

mole ratio of C:Ti= 2:1
4024.98 mols C will yield [1/2*4024.98 mols] Ti(s)

= 2012.49 mols Ti(s) will be produced

Molar mass of Ti= 47.87 g/mol
Mass of Ti produced = Moles of Ti produced *molar mass = 2012.49 mols *47.87 g/mol
=96337.89 g = 96.34 Kg
This is the theoretical yield of Ti(s)
Actual yield = 62.40 kg

% yield = Actual yield/theoretical yield *100
= 62.4o/96.34 *100 = 64.77 %
***************************
Thanks !