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 5.86 l of a gas mixture (at stp) that contains an equal number of carbon monoxide (co) and hydrogen gas (h2) molecules?

2.10 grams. The balanced equation for the reaction is CO + 2H2 ==> CH3OH The key thing to take from this equation is that it takes 2 hydrogen molecules per carbon monoxide molecule for this reaction. And since we've been given an equal number of molecules for each reactant, the limiting reactant will be hydrogen. We can effectively claim that we have 5.86/2 = 2.93 l of hydrogen and an excess of CO to consume all of the hydrogen. So the number of moles of hydrogen gas we have is: 2.93 l / 22.4 l/mol = 0.130803571 mol And since it takes 2 moles of hydrogen gas to make 1 mole of methanol, divide by 2, getting. 0.130803571 mol / 2 = 0.065401786 mol Now we just need to multiply the number of moles of methanol by its molar mass. First lookup the atomic weights involved. Atomic weight carbon = 12.0107 g/mol Atomic weight hydrogen = 1.00794 g/mol Atomic weight oxygen = 15.999 g/mol Molar mass CH3OH = 12.0107 + 4 * 1.00794 + 15.999 = 32.04146 g/mol So the mass produced is 32.04146 g/mol * 0.065401786 mol = 2.095568701 g And of course, properly round the answer to 3 significant digits, giving 2.10 grams.

It starts to rust i think