Respuesta :
Answer:
The percent yield of the reaction is 62.05 %
Explanation:
Step 1: Data given
Volume of methane = 25.5 L
Pressure of methane = 732 torr
Temperature = 25.0 °C = 298 K
Volume of water vapor = 22.0 L
Pressure of H2O = 704 torr
Temperature = 125 °C
The reaction produces 26.0 L of hydrogen gas measured at STP
Step 2: The balanced equation
CH4(g) + H2O(g) → CO(g) + 3H2(g)
Step 3: Calculate moles methane
p*V = n*R*T
⇒with p = the pressure of methane = 0.963158 atm
⇒with V = the volume of methane = 25.5 L
⇒with n = the moles of methane = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 298 K
n = (p*V) / (R*T)
n = (0.963158 * 25.5 ) / ( 0.08206 * 298)
n = 1.0044 moles
Step 4: Calculate moles H2O
p*V = n*R*T
⇒with p = the pressure of methane = 0.926316 atm
⇒with V = the volume of methane = 22.0 L
⇒with n = the moles of methane = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*atm/mol*K
⇒with T = the temperature = 398 K
n = (p*V) / (R*T)
n = (0.926316 * 22.0) / (0.08206 * 398)
n = 0.624 moles
Step 5: Calculate the limiting reactant
For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2
H2O is the limiting reactant. It will completely be consumed (0.624 moles).
Methane is in excess. There will react 0.624 moles. There will remain 1.0044 - 0.624 moles = 0.3804 moles methane
Step 6: Calculate moles hydrogen gas
For 1 mol methane we need 1 mol H2O to produce 1 mol CO and 3 moles H2
For 0.624 moles H2O we'll have 3*0.624 = 1.872 moles
Step 9: Calculate volume of H2 at STP
1.0 mol at STP has a volume of 22.4 L
1.872 moles has a volume of 1.872 * 22.4 = 41.9 L
Step 10: Calculate the percent yield of the reaction
% yield = (actual yield / theoretical yield) * 100 %
% yield = ( 26.0 L / 41.9 L) *100 %
% yield = 62.05 %
The percent yield of the reaction is 62.05 %
Answer:
62.02 %
Explanation:
The percent yield of the reaction can be calculated using the following equation:
[tex] \% = \frac{y_{e}}{y_{t}}*100 [/tex]
Where:
[tex]y_{e}[/tex]: is the experimental yield
[tex]y_{t}[/tex]: is the theoretical yield
We can see that we need to find the theoretical yield and the experimental yield.
To calculate the theoretical yield we need to find the number of moles of the reactants using the Ideal Gas Law:
[tex] n_{CH_{4}} = \frac{PV}{RT} [/tex]
Where:
P: is the pressure of the gas = 732 torr
V: is the volume of the gas = 25.5 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 25 °C
Hence the number of moles of the methane is:
[tex] n_{CH_{4}} = \frac{732 torr \cdot \frac{1 atm}{760 torr}*25.5 L}{0.082 L*atm*K^{-1}*mol^{-1}*(25 + 273 K)} = 1.005 moles [/tex]
Similarly, the number of moles of the water vapor is:
[tex] n_{H_{2}O} = \frac{704 torr \cdot \frac{1 atm}{760 torr}*22.0 L}{0.082 L*atm*K^{-1}*mol^{-1}*(125 + 273 K)} = 0.624 moles [/tex]
Now, we need to find the limiting reactan. In the following equation:
CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)
we have that 1 mol of CH₄ reacts with 1 mol of H₂O:
[tex] \frac{1 mol CH_{4}}{1 mol H_{2}O}*0.624 moles H_{2}O = 0.624 moles CH_{4} [/tex]
We need 0.624 moles of CH₄ to react with H₂O, and we have 1.005 moles of CH₄. Therefore, the limiting reactant is the H₂O.
Since 1 mol of H₂O produces 3 moles of H₂, the number of H₂ moles produced is:
[tex] \frac{3 mol H_{2}}{ 1 mol H_{2}O}*0.624 moles H_{2}O = 1.872 moles [/tex]
The experimental yield is:
[tex]n = \frac{PV}{RT} = \frac{1 atm*26.0 L}{0.082 L*atm*K^{-1}*mol^{-1}*273 K} = 1.161 moles[/tex]
Finally, the percent yield of the reaction is:
[tex] % y = \frac{26.0 L}{41.91 L}* 100 = 62.04 /% [/tex][tex] \% y = \frac{1.161}{1.872}*100 = 62.02 \% [/tex]
I hope it helps you!
