Question

Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction: CO(g)+2H2(g)→CH3OH(g)CO(g)+2H2(g)→CH3OH(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.

Answer 1

Answer: 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:

$PV=nRT$            ..........(1)

• For carbon monoxide:

We are given:

$P=232mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$232mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{232\times 1.55}{62.3637\times 305}=0.0189mol$

• For hydrogen gas:

We are given:

$P=389mmHg\\V=1.55L\\T=305K\\R=62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

Putting values in equation 1, we get:

$389mmHg\times 1.55L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 305K\\\\n=\frac{389\times 1.55}{62.3637\times 305}=0.0317mol$

For the given chemical equation:

$CO(g)+2H_2(g)\rightarrow CH_3OH(g)$

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 = $\frac{1}{2}\times 0.0317=0.01585mol$ 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 = $\frac{1}{2}\times 0.0317=0.01585moles$ 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:

$0.01585mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.01585mol\times 32g/mol)=0.507g$

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