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 : The mass of 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 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:

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 0.294 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, 0.294 mole of react to give mole of

Now we have to calculate the mass of

Molar mass of = 32 g/mole

Therefore, the mass of produced is, 4.704 grams.

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: