When SbCl3(g) (4.867 mol) and 0.01217 mol/L of Cl2(g) in a 400.0 L reaction vessel at 958.0 K are allowed to come to equilibrium the mixture contains 0.008586 mol/L of SbCl5(g). What concentration (mol/L) of SbCl3(g) reacted?
SbCl3(g)+Cl2(g) = SbCl5(g)
Unit Conversions:
K = C + 273
Molar Mass (g/mol)
SbCl3(g) 228.16
Cl2(g) 70.906
SbCl5(g) 299.06
To find the concentration (mol/L) of SbCl3(g) reacted, we can use the stoichiometry of the reaction and the given concentrations.
The balanced equation for the reaction is:
SbCl3(g) + Cl2(g) -> SbCl5(g)
From the balanced equation, we can see that the stoichiometric ratio between SbCl3 and SbCl5 is 1:1. This means that for every 1 mol of SbCl3 reacted, 1 mol of SbCl5 is formed.
First, let's calculate the number of moles of SbCl5(g) formed:
Given concentration of SbCl5(g) = 0.008586 mol/L
Total volume of the reaction vessel = 400.0 L
Number of moles of SbCl5(g) = concentration x volume = 0.008586 mol/L x 400.0 L = 3.4344 mol
Since the stoichiometric ratio is 1:1, this means that 3.4344 mol of SbCl3 reacted.
Now, let's calculate the concentration (mol/L) of SbCl3(g) that reacted:
Total initial moles of SbCl3(g) = 4.867 mol
Initial concentration of SbCl3(g) = total initial moles / total volume = 4.867 mol / 400.0 L = 0.01216875 mol/L
Change in concentration of SbCl3(g) = initial concentration - concentration of SbCl3(g) at equilibrium = 0.01216875 mol/L - 3.4344 mol/400.0 L = -0.008586 mol/L
Since the concentration of a reactant decreases during the reaction, the negative sign indicates that the concentration of SbCl3(g) reacted.
Therefore, the concentration (mol/L) of SbCl3(g) reacted is 0.008586 mol/L (negative value).
To find the concentration (mol/L) of SbCl3(g) that reacted, we need to use the balanced equation for the reaction and the stoichiometry of the reaction.
The balanced equation for the reaction is:
SbCl3(g) + Cl2(g) → SbCl5(g)
According to the problem statement, the initial amount of SbCl3(g) is given as 4.867 mol and the amount of Cl2(g) is given as 0.01217 mol/L. The final concentration of SbCl5(g) is given as 0.008586 mol/L.
Let's calculate the amount of SbCl3(g) that reacted.
First, we calculate the initial moles of SbCl3(g) by multiplying the initial concentration (0.01217 mol/L) by the volume (400.0 L):
Initial moles of SbCl3(g) = (0.01217 mol/L) x (400.0 L) = 4.868 mol
Now, we can use the stoichiometry of the reaction to find the amount of SbCl3(g) that reacted:
Moles of SbCl3(g) reacted = Initial moles of SbCl3(g) - Moles of SbCl3(g) remaining in equilibrium
The moles of SbCl3(g) remaining in equilibrium can be calculated by using the final concentration and volume:
Moles of SbCl3(g) remaining = (0.008586 mol/L) x (400.0 L) = 3.4344 mol
Now, we can find the moles of SbCl3(g) that reacted:
Moles of SbCl3(g) reacted = Initial moles of SbCl3(g) - Moles of SbCl3(g) remaining = 4.868 mol - 3.4344 mol = 1.4336 mol
Finally, we can calculate the concentration of SbCl3(g) that reacted by dividing the moles of SbCl3(g) reacted by the final volume:
Concentration of SbCl3(g) reacted = (1.4336 mol) / (400.0 L) = 0.003584 mol/L
Therefore, the concentration of SbCl3(g) that reacted is 0.003584 mol/L.