a chemist carries out the reaction of 0.250 mol of f2 with excess of cl2. At 250 degress C and 1.00 atm, what is the maximum volume of CLF3 in liters that the reaction can produce?
3F2 + 3Cl2 ==> 2ClF3
mols F2 = 0.250
mols ClF3 = 0.250 x (2 mols ClF3/3 mols F2) = 0.167 mols F2.
Convert this to volume using PV = nRT
PV=nRT
(1)(x)= (.250)(0.0821)(523)
1x= 10.73
Actually, the answer is 7.16 L
To determine the maximum volume of CLF3 that can be produced in this reaction, we need to use stoichiometry to find the balanced equation, and then use the ideal gas law to calculate the volume.
1. Balanced Equation:
The reaction between F2 and Cl2 results in the formation of CLF3. To write the balanced equation, we start with the reactants and determine the stoichiometric coefficients.
F2 + Cl2 -> 2CLF3
2. Calculate the limiting reactant:
Since the given quantity of F2 is 0.250 mol and it is reacted with an excess of Cl2, it means that F2 is the limiting reactant. Therefore, we will use the given quantity of F2 to calculate the amount of product formed.
3. Convert moles of F2 to moles of CLF3:
According to the balanced equation, 1 mole of F2 reacts to form 2 moles of CLF3. Therefore, the 0.250 mol of F2 will produce 0.250 mol x 2 moles CLF3/mol = 0.500 mol of CLF3.
4. Convert moles of CLF3 to volume:
To convert moles of CLF3 to volume, we will use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Given:
n = 0.500 mol (calculated in step 3)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 250°C = 250 + 273 = 523 K (converted to Kelvin)
5. Solve for volume:
We can rearrange the ideal gas law equation to solve for V:
V = nRT/P
V = (0.500 mol) x (0.0821 L·atm/(mol·K)) x (523 K) / (1.00 atm)
Calculating the expression, we find:
V ≈ 21.32 L
Therefore, the maximum volume of CLF3 that the reaction can produce is approximately 21.32 liters.