Carbon monoxide gas reacts with hydrogen gas to form methanol: CO (g_ + 2H2 (g) → CH3OH (g) A 1.50L 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 397 mmHg. Identify the limiting reactant and determine the theoretical yield of methanol in grams.

: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 .

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:

and,

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:

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of  react with 1 mole of  

So, 0.0601 moles of  react with  moles of  

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  

From the reaction, we conclude that

As, 2 mole of  react to give 1 mole of  

So, 0.0601 moles of  react with  moles of  

Now we have to calculate the mass of  

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