Answer : The limiting reactant is
and the theoretical yield of methanol is, 0.96 grams.
Explanation :
First we have to calculate the moles of
and
.
![P_{CO}V=n_{CO}RT](https://tex.z-dn.net/?f=P_%7BCO%7DV%3Dn_%7BCO%7DRT)
where,
= pressure of CO gas = 232 mmHg = 0.305 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of CO gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
![(0.305atm)\times (3.06L)=n_{CO}\times (0.0821L.atm/mol.K)\times (305K)](https://tex.z-dn.net/?f=%280.305atm%29%5Ctimes%20%283.06L%29%3Dn_%7BCO%7D%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28305K%29)
![n_{CO}=0.0373mol](https://tex.z-dn.net/?f=n_%7BCO%7D%3D0.0373mol)
and,
![P_{H_2}V=n_{H_2}RT](https://tex.z-dn.net/?f=P_%7BH_2%7DV%3Dn_%7BH_2%7DRT)
where,
= pressure of
gas = 374 mmHg = 0.492 atm (1 atm = 760 mmHg)
V = volume of gas = 1.65 L
T = temperature of gas = 305 K
= number of moles of
gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
![(0.492atm)\times (3.06L)=n_{H_2}\times (0.0821L.atm/mol.K)\times (305K)](https://tex.z-dn.net/?f=%280.492atm%29%5Ctimes%20%283.06L%29%3Dn_%7BH_2%7D%5Ctimes%20%280.0821L.atm%2Fmol.K%29%5Ctimes%20%28305K%29)
![n_{H_2}=0.0601mol](https://tex.z-dn.net/?f=n_%7BH_2%7D%3D0.0601mol)
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
![CO(g)+2H_2(g)\rightarrow CH_3OH(g)](https://tex.z-dn.net/?f=CO%28g%29%2B2H_2%28g%29%5Crightarrow%20CH_3OH%28g%29)
From the balanced reaction we conclude that
As, 2 mole of
react with 1 mole of ![CO](https://tex.z-dn.net/?f=CO)
So, 0.0601 moles of
react with
moles of ![CO](https://tex.z-dn.net/?f=CO)
From this we conclude that,
is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of ![CH_3OH](https://tex.z-dn.net/?f=CH_3OH)
From the reaction, we conclude that
As, 2 mole of
react to give 1 mole of ![CH_3OH](https://tex.z-dn.net/?f=CH_3OH)
So, 0.0601 moles of
react with
moles of ![CH_3OH](https://tex.z-dn.net/?f=CH_3OH)
Now we have to calculate the mass of ![CH_3OH](https://tex.z-dn.net/?f=CH_3OH)
![\text{ Mass of }CH_3OH=\text{ Moles of }CH_3OH\times \text{ Molar mass of }CH_3OH](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DCH_3OH%3D%5Ctext%7B%20Moles%20of%20%7DCH_3OH%5Ctimes%20%5Ctext%7B%20Molar%20mass%20of%20%7DCH_3OH)
![\text{ Mass of }CH_3OH=(0.0300moles)\times (32g/mole)=0.96g](https://tex.z-dn.net/?f=%5Ctext%7B%20Mass%20of%20%7DCH_3OH%3D%280.0300moles%29%5Ctimes%20%2832g%2Fmole%29%3D0.96g)
Therefore, the theoretical yield of methanol is, 0.96 grams.