I need help on some of the steps in this problem about gases.
An 8.55 mol sample of methanol,CH_3OH, is placed in a 25L evacuated rigid tank and heated to 427C. At that temperature, all of the menthanol is vaporized and some of the menthanol decomposes to form carbon monoxide gas and hydrogen gas, as represented in the equation below.
CH_3OH(g)----CO(g) + 2H_2

Question

I need help on some of the steps in this problem about gases.
An 8.55 mol sample of methanol,CH_3OH, is placed in a 25L evacuated rigid tank and heated to 427C. At that temperature, all of the menthanol is vaporized and some of the menthanol decomposes to form carbon monoxide gas and hydrogen gas, as represented in the equation below.
           CH_3OH(g)---->CO(g) + 2H_2

anonymous · Answer

the first part is
a)the reaction mixture contains 6.30 mol of CO9g) at the end at 327C
i.) calculate the number of moles of H_2(g) in the tank

i figured since the ratio of CO to h_2 is 1:2 then[ 6.30mol of CO x 2=12.6 mol of H_2].

is that correct?

anonymous · Answer

@timo86m

anonymous · Answer

@dan815  @iambatman

anonymous · Answer

hi there,,,, from the equation above methanol gives 2 mole of H2

anonymous · Answer

one methanol gives 2 mole of h2

anonymous · Answer

8.55*2=17.1 mole of H2

jfraser · Answer

the methanol doesn't COMPLETELY decompose, so even though there starts with 8.55 moles of methanol, only 6.30 moles decomposes so 6.30 moles of CO also means 12.6 moles of H2. You are correct