Respuesta :
Answer: The limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams
Explanation:
To calculate the number of moles, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex] ..........(1)
- For carbon monoxide:
We are given:
[tex]P=232mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
Putting values in equation 1, we get:
[tex]232mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{232\times 1.55}{62.3637\times 305}=0.0189mol[/tex]
- For hydrogen gas:
We are given:
[tex]P=389mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
Putting values in equation 1, we get:
[tex]389mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{389\times 1.55}{62.3637\times 305}=0.0317mol[/tex]
For the given chemical equation:
[tex]CO(g)+2H_2(g)\rightarrow CH_3OH(g)[/tex]
By Stoichiometry of the reaction:
2 moles of hydrogen gas reacts with 1 mole of carbon monoxide
So, 0.0317 moles of hydrogen gas will react with = [tex]\frac{1}{2}\times 0.0317=0.01585mol[/tex] of carbon monoxide
As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.
Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of hydrogen gas produces 1 mole of methanol
So, 0.0317 moles of hydrogen gas will produce = [tex]\frac{1}{2}\times 0.0317=0.01585moles[/tex] of methanol
Now, calculating the mass of methanol from equation 1, we get:
Molar mass of methanol = 32 g/mol
Moles of methanol = 0.01585 moles
Putting values in equation 1, we get:
[tex]0.01585mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.01585mol\times 32g/mol)=0.507g[/tex]
Hence, the limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams
