Carbon monoxide, CO, and hydrogen, H2, react according to the reaction below.
4 CO(g)+ 9 H2(g)--> C4H10(g)+ 4 H2O(g)
What volume of the excess reactant remains if 19.6 L CO and 27.2 L H2 are allowed to react. Assume that the volumes of both gases are measured at 713oC and 1.35 atm.
Give your answer in litres, accurate to two significant figures. Do not include the units as part of your answer.
To solve this problem, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To do this, we can use the concept of stoichiometry and compare the number of moles of each reactant to their respective coefficients in the balanced chemical equation.
Let's start by calculating the number of moles of CO and H2, using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.35 atm)
V = volume (19.6 L for CO and 27.2 L for H2)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (convert 713°C to Kelvin: 713 + 273 = 986 K)
For CO:
n_CO = PV / RT = (1.35 atm)(19.6 L) / (0.0821 L·atm/mol·K)(986 K)
For H2:
n_H2 = PV / RT = (1.35 atm)(27.2 L) / (0.0821 L·atm/mol·K)(986 K)
Now, we can compare the moles of CO and H2 to their coefficients in the balanced equation:
Moles of CO: n_CO = (1.35 atm)(19.6 L) / (0.0821 L·atm/mol·K)(986 K)
Moles of H2: n_H2 = (1.35 atm)(27.2 L) / (0.0821 L·atm/mol·K)(986 K)
Next, we calculate the moles of the product formed (C4H10) using the stoichiometric ratio:
4 moles of CO = 1 mole of C4H10
9 moles of H2 = 1 mole of C4H10
We need to determine which reactant produces the smaller amount of C4H10. We can calculate the theoretical moles of C4H10 produced:
Theoretical moles of C4H10 from CO = n_CO / 4
Theoretical moles of C4H10 from H2 = n_H2 / 9
Finally, we compare the theoretical moles of C4H10 produced from each reactant to find the limiting reactant:
The limiting reactant will be the one that produces the smaller amount of C4H10.
Once we have determined the limiting reactant, we can calculate the moles of the excess reactant that remains by subtracting the moles of the limiting reactant from the respective initial moles of the reactant.
Finally, we use the ideal gas law again to convert the moles of the excess reactant to the volume at the given conditions. Since the pressure and temperature are the same for both CO and H2, the volume of the excess reactant will be in the same proportion as the moles.
Keep in mind that we are assuming ideal gas behavior throughout these calculations.