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 : 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.

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