Carbon monoxide, CO and hydrogen, H2, react accordingly:
4CO(g)+9H2(g)-->C4H10(g)+4H2O
What volume of the EXCESS REACTANT remains if 17.2L CO and 42.1L H2 are allowed to react?
Assume both gases are measured at 711 degree celsius and 1.33atm.
This is a limiting ragent problem. I know that because amounts are given for BOTH reactants.
When dealing with gases and the T and P remain the same, one can take a short cut ans use volume directly as if volume = mols. It saves a little time.
a. Calculate volume butane formed with 17.2 L CO and all the H2 needed. That's 17.2 L CO x (1 mol C4H10/4 mols CO) = 17.2/4 = approx 4.30 L.
b. Calculate volume butane formed with 42.1L H2 and all the CO needed. That's
42.1 L H2 x (1 mol C4H10/9 mols H2) = approx 4.7.
c. The values for L C4H10 don't agree which means one is wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.
d. That means CO is the limiting reagent and H2 is the excess reagent.
e. How much H2 will be used up with the CO? That's 17.2 L CO x (9 mol H2/4 mol CO) = about 38.7 L H2 used.
f. We started with 42.1 LH2; we used 38.7 L H2, the difference is the amount H2 not used.
To determine the excess reactant remaining, we first need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed first and determines the amount of product formed.
1. Convert the volumes of CO and H2 to the number of moles using the ideal gas law equation: PV = nRT.
For CO:
P = 1.33 atm
V = 17.2 L
T = 711 °C + 273.15 = 984.15 K (convert to Kelvin)
R = 0.0821 L.atm/mol.K
n_CO = (P * V) / (R * T)
= (1.33 atm * 17.2 L) / (0.0821 L.atm/mol.K * 984.15 K)
≈ 0.99 moles
For H2:
P = 1.33 atm
V = 42.1 L
T = 984.15 K
R = 0.0821 L.atm/mol.K
n_H2 = (P * V) / (R * T)
= (1.33 atm * 42.1 L) / (0.0821 L.atm/mol.K * 984.15 K)
≈ 2.07 moles
2. Calculate the stoichiometric ratio of moles of CO to moles of H2 using the balanced equation:
4CO(g) + 9H2(g) → C4H10(g) + 4H2O(g)
Based on the balanced equation, the stoichiometric ratio is:
4 moles CO : 9 moles H2
3. Determine the limiting reactant based on the stoichiometric ratio.
To do this, divide the number of moles of each reactant by its coefficient in the balanced equation:
For CO: 0.99 moles CO / 4 = 0.24875
For H2: 2.07 moles H2 / 9 = 0.23
Since the ratio for CO is larger, it is the limiting reactant.
4. Determine the moles of C4H10 produced.
Using the stoichiometric ratio, we find that for every 4 moles of CO, we produce 1 mole of C4H10. Therefore,
n_C4H10 = (0.99 moles CO) / (4 moles CO/1 mole C4H10)
≈ 0.2475 moles C4H10
5. Calculate the volume of excess reactant remaining.
Now that we know the moles of C4H10 produced, we can determine the moles of H2 consumed in the reaction using the stoichiometric ratio:
moles H2 consumed = (9 moles H2/4 moles CO) * (0.99 moles CO)
≈ 2.2275 moles H2
moles H2 remaining = 2.07 moles H2 - 2.2275 moles H2
= -0.1575 moles H2 (negative because H2 is in excess)
To convert moles of H2 remaining to volume at the given conditions (711 °C and 1.33 atm), use the ideal gas law equation:
PV = nRT
P = 1.33 atm
n_H2 = -0.1575 moles H2
R = 0.0821 L.atm/mol.K
T = 984.15 K
V_H2 = (n_H2 * R * T) / P
= (-0.1575 moles * 0.0821 L.atm/mol.K * 984.15 K) / 1.33 atm
≈ -8.8 L (negative because H2 is in excess)
Since we can't have a negative volume, we can conclude that there is no excess H2 remaining.