Solid titanium oxide (TiO2) reacts with solid carbon (C) to produce solid titanium (Ti) and gaseous carbon monoxide (CO). When 14.0 kg of C is allowed to react with 44.0 kg of TiO2 7.5 kg of Ti is produced. What is the theoretical yield of Ti expressed in mass and the percent yield?
To find the theoretical yield of Ti and the percent yield, we need to calculate the stoichiometric ratio and then use it to find the expected mass of Ti based on the mass of TiO2.
1. Calculate the stoichiometric ratio between TiO2 and Ti:
Since the balanced chemical equation is not given, we can derive it by examining the reactants and products:
TiO2 + C → Ti + CO
From this equation, we can see that 1 mole of TiO2 reacts with 1 mole of C to produce 1 mole of Ti. Therefore, the stoichiometric ratio between TiO2 and Ti is 1:1.
2. Determine the number of moles of TiO2 used:
Mass of TiO2 = 44.0 kg
Molar mass of TiO2 = 79.866 g/mol
Number of moles of TiO2 = (Mass of TiO2 / Molar mass of TiO2)
= (44,000 g / 79.866 g/mol)
= 550.66 mol
3. Calculate the expected mass of Ti based on the stoichiometric ratio:
Since the stoichiometric ratio is 1:1 between TiO2 and Ti, the number of moles of Ti will be the same as the number of moles of TiO2.
Number of moles of Ti = 550.66 mol
Expected mass of Ti = (Number of moles of Ti x Molar mass of Ti)
= (550.66 mol x 47.867 g/mol)
= 26,352.8 g
= 26.35 kg
Therefore, the theoretical yield of Ti is 26.35 kg.
4. Calculate the percent yield:
Actual mass of Ti = 7.5 kg
Percent yield = (Actual mass of Ti / Theoretical mass of Ti) x 100
= (7.5 kg / 26.35 kg) x 100
= 28.46%
Therefore, the percent yield of Ti is 28.46%.
To determine the theoretical yield of Ti (mass of Ti produced) and the percent yield, we need to use stoichiometry and balance the chemical equation.
The balanced chemical equation for the reaction between TiO2 and C is:
TiO2 + 2C → Ti + 2CO
From the balanced equation, we can see that 1 mole of TiO2 reacts with 2 moles of C to produce 1 mole of Ti.
Step 1: Convert the masses of C and TiO2 to moles.
Using the molar mass of each substance, we can convert the masses to moles.
Molar mass of C: 12.01 g/mol
Molar mass of TiO2: 79.87 g/mol
Number of moles of C = mass of C / molar mass of C
Number of moles of C = 14.0 kg * 1000 g/kg / 12.01 g/mol ≈ 1162.03 mol
Number of moles of TiO2 = mass of TiO2 / molar mass of TiO2
Number of moles of TiO2 = 44.0 kg * 1000 g/kg / 79.87 g/mol ≈ 550.34 mol
Step 2: Determine the limiting reactant.
To determine the limiting reactant, compare the number of moles of each reactant in the balanced equation.
From the balanced equation, it is clear that 1 mole of TiO2 reacts with 2 moles of C. Therefore, we need twice as many moles of C as moles of TiO2.
Since we have 1162.03 moles of C and 550.34 moles of TiO2, TiO2 is the limiting reactant. The reaction will consume all the TiO2 and leave excess carbon.
Step 3: Calculate the theoretical yield of Ti.
From the balanced equation, 1 mole of TiO2 produces 1 mole of Ti.
Therefore, moles of Ti = moles of TiO2 = 550.34 mol.
Mass of Ti = moles of Ti * molar mass of Ti
Mass of Ti = 550.34 mol * 47.87 g/mol ≈ 26,352.67 g ≈ 26.35 kg
The theoretical yield of Ti is approximately 26.35 kg.
Step 4: Calculate the percent yield.
Percent yield is calculated by dividing the actual yield (7.5 kg) by the theoretical yield (26.35 kg) and multiplying by 100.
Percent yield = (actual yield / theoretical yield) * 100
Percent yield = (7.5 kg / 26.35 kg) * 100 ≈ 28.47%
The percent yield is approximately 28.47%.
TiO2 + 2C ==> 2CO + Ti
We don't need to make any corrections for molar mass and kg if we keep everything in kg.
kg mols C = 14.0/12 = approx 1.17
kg mols TiO2 = 44.0/79.88 = approx 0.5
Using the coefficients in the balanced equation, convert kg mols of each to kg mols of Ti.
for C that's 1 x (1 mol Ti/2 mol C) = approx 1.17 x 1/2 = approx 0.59 kg mol Ti from C.
for TiO2 that's 0.5 x (1 mol Ti/1 mol TiO2) = approx 0.5 kg mol Ti.
Note the values for Ti are different; in limiting reagent problems the smaller number is ALWAYS the correct value and the reagent producing that number is the limiting reagent. So we can obtain approx 0.5 kg mol Ti from the TiO2 reaction.
g TiO2 = mols x molar mass = ? kg Ti. This is the theoretical yield (100%).
The actual yield was 7.5kg.
%yield = (actual yield/theoretical yield)*100 = ?
Note I have used approximations. You need to redo all of the numbers.