Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)+2H2(g)→CH3OH(g)CO(g)+2H2(g)→C H3OH(g) A 1.55 LL reaction vessel, initially at 305 KK, contains carbon monoxide gas at a partial pressure of 232 mmHgmmHg and hydrogen gas at a partial pressure of 389 mmHgmmHg . Identify the limiting reactant and determine the theoretical yeild of methonal in grams.

The limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

           ..........(1)

We are given:

Putting values in equation 1, we get:

We are given:

Putting values in equation 1, we get:

For the given chemical equation:

By Stoichiometry of the reaction:

2 moles of hydrogen gas reacts with 1 mole of carbon monoxide

So, 0.0317 moles of hydrogen gas will react with = of carbon monoxide

As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.

Thus, hydrogen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of hydrogen gas produces 1 mole of methanol

So, 0.0317 moles of hydrogen gas will produce = of methanol

Now, calculating the mass of methanol from equation 1, we get:

Molar mass of methanol = 32 g/mol

Moles of methanol = 0.01585 moles

Putting values in equation 1, we get:

Hence, the limiting reactant is hydrogen gas and the theoretical yield of methanol is 0.507 grams