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

Explanation :

First we have to calculate the moles of and gas.

Using ideal gas equation :

where,

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

conversion used : (1 atm = 760 mmHg)

V = Volume of gas = 1.30 L

n = number of moles   gas = ?

R = Gas constant =

T = Temperature of gas = 305 K

Putting values in above equation, we get:

and,

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

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, 0.0264 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, 1 mole of react to give 2 mole of

So, 0.0264 moles of react to give moles of

Now we have to calculate the mass of

Molar mass of methanol = 32 g/mole

Therefore, the theoretical yield of is 0.422 grams.

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