How many grams of gaseous required?
If 0.260 mol of solid C and 20.0 g of solid TiO2 are reacted stoichiometrically according to the balanced equation, how many grams of gaseous Cl2 are required?
3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(l) + 2CO2(g) + 2CO(g)
1. Make sure the equation is balanced.
2a. Convert 0.260 g C to moles. moles = grams/molar mass.
2b Convert 20.0 g TiO2 to moles using the same procedure.
3a. Using the coefficients in the balanced equation, convert moles C in 2a to moles Cl2.
3b. Using the same procedure, convert moles TiO2 in 2b to moles Cl2.
3c. More than likely the answer for 2a and 2b will be different. Obviously both can't be correct; the correct one in limiting reagent problems is ALWAYS the smaller value. (The material that produces the smaller value is the limiting reagent.)
4. Convert the value from 3c to grams. grams = moles Cl2 x molar mass Cl2.
To determine the number of grams of gaseous Cl2 required, you need to use stoichiometry. Stoichiometry is a mathematical relationship between the balanced equation and the amounts of substances involved in a chemical reaction.
First, let's calculate the number of moles of Cl2 required using the given information. We have 0.260 mol of solid C and 20.0 g of solid TiO2. We need to find the limiting reactant to determine the moles of Cl2 required.
Step 1: Calculate the moles of C:
moles of C = mass / molar mass
moles of C = 0.260 mol
Step 2: Calculate the moles of TiO2:
moles of TiO2 = mass / molar mass
molar mass of TiO2 = atomic mass of Ti + 2 * atomic mass of O
molar mass of TiO2 = 47.867 g/mol + (2 * 16.00 g/mol)
molar mass of TiO2 = 79.866 g/mol
moles of TiO2 = 20.0 g / 79.866 g/mol
moles of TiO2 ≈ 0.250 mol
Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the moles of each reactant to their respective stoichiometric coefficients in the balanced equation. The reactant with the smaller number of moles is the limiting reactant, as it restricts the amount of product that can be formed.
From the balanced equation, we see that the stoichiometric coefficient of C is 4, while the stoichiometric coefficient of TiO2 is 6. Since 0.260 mol of C < 0.250 mol of TiO2, C is the limiting reactant.
Step 4: Calculate the moles of Cl2 using the limiting reactant:
From the balanced equation, you see that the stoichiometric coefficient of Cl2 is 6. Thus, for every 6 moles of Cl2, 4 moles of C are required.
moles of Cl2 = (moles of C / stoichiometric coefficient of C) * stoichiometric coefficient of Cl2
moles of Cl2 = (0.260 mol / 4) * 6
moles of Cl2 ≈ 0.390 mol
Step 5: Calculate the grams of Cl2:
To convert moles to grams, use the molar mass of Cl2.
grams of Cl2 = moles of Cl2 * molar mass of Cl2
grams of Cl2 = 0.390 mol * (2 * 35.45 g/mol)
grams of Cl2 ≈ 27.6 g
Therefore, approximately 27.6 grams of gaseous Cl2 are required for the reaction.