I'm confused if the required mass from a two step reaction can be found using moles, grams or liters. This happened when attempting to solve a gas reaction with a solid.

The problem is as follows:

Carbon reacts at higher temperatures with water vapor producing carbon monoxide and hydrogen. Carbon monoxide obtained reacts with water vapor producing carbon dioxide and hydrogen. If it is desired to obtain [itex]89.6\,L[/itex] of hydrogen ([itex]H_{2}[/itex]) measured at standard conditions ([itex]0^{\circ}C[/itex] and [itex]1\,atm[/itex]). Find the mass of carbon if the water vapor has an excess of [itex]50\%[/itex].

The alternatives given are as follows:

[itex]\begin{array}{ll}

1.&10\,g\\

2.&14\,g\\

3.&24\,g\\

4.&36\,g\\

5.&108\,g\\

\end{array}[/itex]

What i thought to assess this problem was to go backwards to find the mass of carbon: (For the sake of brevity I'm omitting the states of matter)

Initially it is happening this:

[itex]C+H_{2}O\rightarrow CO + H_{2}[/itex]

Then it happens this:

[itex]CO+H_{2}O\rightarrow CO_{2}+H_{2}O[/itex]

As it is presented both equations are fully balanced.

The combined process can be summarized as follows:

[itex]C+2H_2O \rightarrow CO_{2}+2H_{2}[/itex]

I thought that this latter equation it can be used:

[itex]C+2H_2O \rightarrow CO_{2}+2H_{2}[/itex]

I assumed that since it is mentioned that [itex]50\%[/itex] of water vapor is used then it would meant that in the whole process there's a total of [itex]1.5y[/itex] grams of that water vapor.

As it is mentioned that [itex]89.6\,L[/itex] is to be obtained, using this information I can obtain the moles of water vapor. Then those moles can be used to obtain the moles of carbon and using the atomic mass of carbon I can obtain the required mass.

Therefore

[itex]\begin{array}{llll}

C&+2H_2O&\rightarrow&CO_{2}&+2H_{2}\\

x&1.5y\times\frac{1\,mol}{18\,g}&&0.75y&+1.5y\\

\end{array}[/itex]

[itex]x=moles of carbon[/itex]

[itex]1.5y\,g\frac{\,mol\,H_{2}O}{18\,g}\times\frac{2\,mol\,H_{2}}{2\,mol\,H_{2}O}=89.6\,L\times\frac{2\,g\,H_{2}}{22.4\,L}[/itex]

[itex]y=96\,g\,H_{2}O[/itex]

Then these grams can be used to obtain the grams of Carbon:

[itex]96\,g\,H_{2}O\times\frac{12\,g\,C}{2\times 18\,g\,H_{2}O}=32\,g\,C[/itex]

Therefore the mass of Carbon will be:

[itex]32\,g.[/itex]

But I'm confused on why it does not appear in any of the alternatives. Why am I obtain this discrepancy?.

Doing the same process yields half of that value [itex]16\,g[/itex]. which has left me confused on why?.

Assuming this:

[itex]1.5y\,mol\,H_{2}O\times\frac{2\,mol\,H_{2}}{2\,mol\,H_{2}O}=89.6\,L\times\frac{1\,mol\,H_{2}}{22.4\,L}[/itex]

[itex]y=2.6667[/itex]

Then using this number of moles of water:

[itex]2.6667\times\frac{1\,mol\,C}{2\,mol\,H_{2}O}\times\frac{12\,g}{1\,mol\,C}=16.0\,g[/itex]

Could it be that I overlooked something?. Should it had been the same if I had considered the volume of the water vapor?. Can someone guide me exactly where did this went wrong?.