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 7.02 L of a gas mixture (at STP) that contains an equal number of carbon monoxide (CO) and hydrogen gas (H2) molecules?

Express the mass in grams to three significant digits.

Answer 1

The reaction between CO and H2 to give CH3OH is

CO+ 2H2 -------->CH3OH

theoretical moles ratio of CO:H2= 1:2

we know that 1 mole of any gas occupies 22.4 liters

7.02 L of mixture of gases occupy 7.02/224 =0.3133 moles of mixture

Equal no of molecules also refer to equal no of moles of each gas ( since 1 mole of any gas contain 6.023*10²³ molecules)

moles of CO= moles of H2= 0.3133/2= 0.15665 moles

Actual moles ratio of CO:H2= 0.15665:0.15665= 1:1

so H2 is limiting reactant since theoretical requirement of CO:H2 is 1:2 where as actual requirement of CO:H2 =1:1

so all the H2 is consumed and 2 mole of H2 gives 1 mole of CH3OH

0.15665 moles of H2 gives 0.15665/2= 0.078325 moles of CH3OH.

molar mass of CH3OH= 32 g/mole

mass of CH3OH required= moles* molar mass =0.078325*32 gm =2.51 gm of CH3OH.