Question

Based on the ideal gas law, there is a simple equivalency that exists between the amount of gas and the volume it occupies. At standard temperature and pressure (STP; 273.15 K and 1 atm, respectively), one mole of gas occupies 22.4 L of volume. What mass of methanol (CH3OH) could you form if you reacted 6.59 L of a gas mixture (at STP) that contains an equal number of carbon monoxide (CO) and hydrogen gas (H2) molecules?

Answer 1

Answer : The mass of $CH_3OH$ produced is, 4.704 grams.

Explanation :

First we have to calculate the moles of mixture of gas.

As we know that at STP, 1 mole of gas occupies 22.4 L volume of gas.

As, 22.4 L volume of gas mixture present in 1 mole of gas mixture

So, 6.59 L volume of gas mixture present in $\frac{6.59}{22.4}=0.294$ mole of gas mixture

As the gas mixture (at STP) contains an equal number of carbon monoxide and hydrogen gas molecules.

Moles of carbon monoxide = Moles of hydrogen has = 0.294 mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$CO+2H_2\rightarrow CH_3OH$

From the balanced reaction we conclude that

As, 2 mole of $H_2$ react with 1 mole of $CO$

So, 0.294 moles of $H_2$ react with $\frac{0.294}{2}=0.147$ moles of $CO$

From this we conclude that, $CO$ is an excess reagent because the given moles are greater than the required moles and $H_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CH_3OH$

From the reaction, we conclude that

As, 2 mole of $H_2$ react to give 1 mole of $CH_3OH$

So, 0.294 mole of $H_2$ react to give $\frac{0.294}{2}=0.147$ mole of $CH_3OH$

Now we have to calculate the mass of $CH_3OH$

$\text{ Mass of }CH_3OH=\text{ Moles of }CH_3OH\times \text{ Molar mass of }CH_3OH$

Molar mass of $CH_3OH$ = 32 g/mole

$\text{ Mass of }CH_3OH=(0.147moles)\times (32g/mole)=4.704g$

Therefore, the mass of $CH_3OH$ produced is, 4.704 grams.