I'm here at a study group with Aaron and while he's stuck on that question, I'm stuck on this one... I think we're gonna take a break but I'd love if someone could help break this down for me.
Carbon monoxide, CO, and hydrogen, H , react according to the reaction below.
2 CO(g) + 5 H (g) --> C H (g) + 2 H O(g)
What volume of the excess reactant remains if 18.6 L CO and 32.2 L H are allowed to react. Assume that the volumes of both gases
are measured at 735 C and 1.19 atm.
Give your answer in litres, accurate to two significant figures.
first change everything to moles, and then recall that at STP, 1 mole occupies 22.4L of volume.
@Steve
Moles of CO2 and H? once I find the moles do I use PV=nRT --> V=nRT/P?
Can you give me the answer so I can try and work backwards. I'm really lost. you don't need to show the steps now that i have the process, I'll get there i just want to know if i get the answer right
To find out what volume of the excess reactant remains, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed during the reaction and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles for each reactant using the ideal gas law equation:
PV = nRT
Where:
P = pressure (1.19 atm)
V = volume (18.6 L for CO and 32.2 L for H)
n = number of moles (what we want to find)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (735 C = 1008 K)
For CO:
n_CO = PV / RT = (1.19 atm * 18.6 L) / (0.0821 L.atm/mol.K * 1008 K)
For H:
n_H = PV / RT = (1.19 atm * 32.2 L) / (0.0821 L.atm/mol.K * 1008 K)
Now we need to determine the stoichiometric ratio between CO and H to find the limiting reactant. According to the balanced equation:
2 CO(g) + 5 H (g) --> C H (g) + 2 H O(g)
The ratio between CO to H is 2:5. This means that for every 2 moles of CO, we need 5 moles of H.
Next, we compare the moles of each reactant to the stoichiometric ratio. If the moles of one reactant are less than what is required by the stoichiometric ratio, then that reactant is the limiting reactant.
Let's calculate the moles of each reactant:
moles_CO = n_CO * 2 (to account for the stoichiometry)
moles_H = n_H * 5 (to account for the stoichiometry)
If moles_CO < moles_H, then CO is the limiting reactant. If moles_H < moles_CO, then H is the limiting reactant.
Once you figure out which reactant is the limiting reactant, you can calculate the amount of excess reactant used and the remaining volume of the excess reactant.
Let me know if you need further assistance with these calculations.