Question

an 8.55mol saple of methanol, CH3OH, is placed in a 15.0L evacuated rigid tank and heated to 327 degrees celsisus. at temperature, all of the methanol is vaporized and some of the methanol decomposes to form carbon monoxide gas and hydrogen gas as represented in the equation CH3OH(g)->CO(g)+2H2(g) (a)the reaction mixture contains 6.30 mol of CO(g) at the equilibrium at 327 degrees celsisus (i)calculate the number of moles of H2(g) in the tank. (ii) calculate the number of grams of CH3OH(g) remaining in the tank. (iii) calculate the mole fraction of H2(g) in the tank.

Answer 1

i) From the equation we can see that there are twice as many moles of hyrogen as there are carbon monoxide. You are given moles of monoxide so just multiply by two. Since you know the moles of all substnances you can find remaining grams by the following: \[6.3molesCO(28molarmassCO)=176.4\]\[12.6molesH_2(2.008molarmassH_2)=25.3\]\[8.55molesCH_3OH(32molarmassMethanol)=273.6\]\[273.6- 176.4+25.3=71.9gMethanol\] mol fraction is just the moles of hydrogen gas over the total moles. \[{12.6molesH_2 \over 12.6+6.3+8.55}\times 100 = 45.9 %\] I believe this is all right.

Answer 2

Thank you.

Answer 3

No prob.