If 0.760 mol of solid TiO2 and 4.20 g of solid C are reacted stoichiometrically according to the balanced equation, how many liters of gaseous CO measured at STP are produced?
3TiO2(s) + 4C(s) + 6Cl2(g) ¡æ 3TiCl4(l) + 2CO2(g) + 2CO(g)
Molar Mass (g/mol)
TiO2 79.878
C 12.011
CO 28.010
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
To answer this question, we need to use stoichiometry and the given information about the amounts of TiO2 and C to calculate the number of moles of CO gas produced. Then, we can use the molar volume at STP to convert the moles of CO gas to liters.
Step 1: Calculate moles of CO produced
From the balanced equation, we can see that the molar ratio between TiO2 and CO is 3:2. Therefore, we need to determine the number of moles of TiO2 reacted and then use the stoichiometric ratio to find the number of moles of CO produced.
Moles of TiO2 = 0.760 mol (given)
According to the stoichiometric ratio, 3 moles of TiO2 produce 2 moles of CO.
So, using this ratio, we can calculate the moles of CO produced:
Moles of CO = (2/3) * Moles of TiO2
Moles of CO = (2/3) * 0.760 mol
Step 2: Convert moles of CO to liters of CO gas
Now that we have the number of moles of CO gas produced, we can use the molar volume at STP to convert it to liters.
Molar volume at STP = 22.4 L/mol (given)
Moles of CO2 = (2/3) * 0.760 mol
Liters of CO = Moles of CO * Molar volume at STP
Liters of CO = [(2/3) * 0.760 mol] * 22.4 L/mol
Using this equation, calculate the value for the number of liters of gaseous CO produced.
To find the volume of gaseous CO produced, we first need to calculate the number of moles of CO produced using the given amount of reactants.
1. Calculate the number of moles of TiO2:
moles of TiO2 = mass of TiO2 / molar mass of TiO2
moles of TiO2 = 0.760 mol
2. Calculate the number of moles of C:
moles of C = mass of C / molar mass of C
moles of C = 4.20 g / 12.011 g/mol
moles of C = 0.3498 mol
3. Determine the limiting reactant:
Since the stoichiometric ratio between TiO2 and C is 3:4, we can compare the moles of each reactant to see which is limiting. The reactant that produces fewer moles of products is the limiting reactant.
The moles of CO produced from TiO2 is 2/3 * moles of TiO2 = 2/3 * 0.760 mol = 0.5067 mol
The moles of CO produced from C is 1/4 * moles of C = 1/4 * 0.3498 mol = 0.08745 mol
Since the moles of CO produced from C is smaller, C is the limiting reactant.
4. Calculate the moles of CO produced:
moles of CO = 1/4 * moles of C (from step 3)
moles of CO = 1/4 * 0.3498 mol = 0.08745 mol
5. Convert moles of CO to volume at STP:
volume of CO = moles of CO * molar volume at STP
volume of CO = 0.08745 mol * 22.4 L/mol
volume of CO = 1.9596 L
Therefore, approximately 1.9596 liters of gaseous CO are produced.