Question

Carbon monoxide gas reacts with hydrogen gas to form methanol.CO(g)+2H2(g)?CH3OH(g)A 1.30L 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 387mmHg .Identify the limiting reactant and determine the theoretical yield of methanol in grams.

Answer 1

Answer : The limiting reagent is hydrogen gas and the theoretical yield of $CH_3OH$ is 0.422 grams.

Explanation :

First we have to calculate the moles of $CO_2$ and $H_2$ gas.

Using ideal gas equation :

$PV=nRT\\\\n_{CO_2}=\frac{P_{CO_2}V}{RT}$

where,

P = partial pressure of $CO_2$ gas = 232 mmHg = 0.305 atm

conversion used : (1 atm = 760 mmHg)

V = Volume of gas = 1.30 L

n = number of moles  $CO_2$ gas = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of gas = 305 K

Putting values in above equation, we get:

$n_{CO_2}=\frac{(0.305atm)\times (1.30L)}{(0.0821L.atm/mol.K)\times (305K)}$

$n_{CO_2}=0.0158mole$

and,

P = partial pressure of $H_2$ gas = 387 mmHg = 0.509 atm

$n_{H_2}=\frac{(0.509atm)\times (1.30L)}{(0.0821L.atm/mol.K)\times (305K)}$

$n_{H_2}=0.0264mole$

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, 0.0264 moles of $H_2$ react with $\frac{0.0264}{2}=0.0132$ 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, 1 mole of $H_2$ react to give 2 mole of $CH_3OH$

So, 0.0264 moles of $H_2$ react to give $\frac{0.0264}{2}=0.0132$ moles of $CH_3OH$

Now we have to calculate the mass of $CH_3OH$

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

Molar mass of methanol = 32 g/mole

$\text{ Mass of }CH_3OH=(0.0132moles)\times (32g/mole)=0.422g$

Therefore, the theoretical yield of $CH_3OH$ is 0.422 grams.