Question

Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)+2H2(g)→CH3OH(g) A 1.65 L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 374 mmHg .Identify the limiting reactant and determine the theoretical yeild of methonal in grams.

Answer 1

Answer : The limiting reactant is $H_2$ and the theoretical yield of methanol is, 0.96 grams.

Explanation :

First we have to calculate the moles of $CO$ and $H_2$.

$P_{CO}V=n_{CO}RT$

where,

$P_{CO}$ = 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

$n_{CO}$ = 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:

$(0.305atm)\times (3.06L)=n_{CO}\times (0.0821L.atm/mol.K)\times (305K)$

$n_{CO}=0.0373mol$

and,

$P_{H_2}V=n_{H_2}RT$

where,

$P_{H_2}$ = pressure of $H_2$ gas = 374 mmHg = 0.492 atm   (1 atm = 760 mmHg)

V = volume of gas = 1.65 L

T = temperature of gas = 305 K

$n_{H_2}$ = number of moles of $H_2$ gas = ?

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(0.492atm)\times (3.06L)=n_{H_2}\times (0.0821L.atm/mol.K)\times (305K)$

$n_{H_2}=0.0601mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

From the balanced reaction we conclude that

As, 2 mole of $H_2$ react with 1 mole of $CO$

So, 0.0601 moles of $H_2$ react with $\frac{0.0601}{2}=0.0300$ moles of $CO$

From this we conclude that, $CO$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CH_3OH$

From the reaction, we conclude that

As, 2 mole of $H_2$ react to give 1 mole of $CH_3OH$

So, 0.0601 moles of $H_2$ react with $\frac{0.0601}{2}=0.0300$ moles of $CH_3OH$

Now we have to calculate the mass of $CH_3OH$

$\text{ Mass of }CH_3OH=\text{ Moles of }CH_3OH\times \text{ Molar mass of }CH_3OH$

$\text{ Mass of }CH_3OH=(0.0300moles)\times (32g/mole)=0.96g$

Therefore, the theoretical yield of methanol is, 0.96 grams.

Answer 2

For the reaction CO(g) + 2H₂(g) → CH₃OH(g), knowing that the initials conditions of volume, temperature, and pressure of CO and H₂ are 1.65 L,  305 K, 232 mmHg, and 374 mmHg, respectively, we have:

1. The limiting reactant is H₂

2. The theoretical yield of methanol is 0.513 grams.

The reaction is:

CO(g) + 2H₂(g) → CH₃OH(g)   (1)

We have:

V: is the volume = 1.65 L

T: is the temperature = 305 K

$P_{CO}$: is the pressure of CO = 232 mmHg

$P_{H_{2}}$: is the pressure of H₂ = 374 mmHg

Calculating the number of moles

Using the ideal gas law we can find the number of moles of the gases (n).

$n = \frac{PV}{RT}$

Where:

R: is the gas constant = 0.082 L*atm/(K*mol)

The initial number of moles of CO and H₂ is:

$n_{CO_{i}} = \frac{P_{CO}V}{RT} = \frac{232 mmHg*\frac{1 atm}{760 mmHg}*1.65 L}{0.082 L*atm/(K*mol)*305 K} = 0.020 moles$

$n_{H_{2}_{i}} = \frac{P_{H_{2}}V}{RT} = \frac{374 mmHg*\frac{1 atm}{760 mmHg}*1.65 L}{0.082 L*atm/(K*mol)*305 K} = 0.032 moles$

Hence, the number of moles of CO and H₂ are 0.020 moles and 0.032 moles, respectively.

Finding the limiting reactant

From the stoichiometric relation between CO and H₂ of reaction (1) we can find the limiting reactant as follows:

$n_{CO} = \frac{1\: mol\: CO}{2\: mol\: H_{2}}*n_{H_{2}_{i}} = \frac{1\: mol\: CO}{2\: mol\: H_{2}}*0.032 \:moles = 0.016\: moles$

Initially, we have 0.020 moles of CO and we can see that 0.016 moles of CO react with H₂, so, the limiting reactant is H₂.    

Determining the theoretical yield

Knowing that the limiting reactant is H₂ and its stoichiometric relation with CH₃OH, we can find the theoretical yield as follows:

$n_{CH_{3}OH} = \frac{1\: mol\: CH_{3}OH}{2\: mol\: H_{2}}*n_{H_{2}}_{i}} = \frac{1\: mol\: CH_{3}OH}{2\: mol\: H_{2}}*0.032 \:moles} = 0.016 \: moles$

Now, the theoretical yield is:

$m_{CH_{3}OH} = n_{CH_{3}OH}*M = 0.016\: moles*32.04 g/mol = 0.513 g$

Therefore, the theoretical yield of methanol is 0.513 g.

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