Calculating required mass of reactant to produce a specific product

I want to know how many grams of $\ce{C}$ (carbon) are required to create $\pu{58 grams}$ of $\ce{Na2S}$, given
$$\ce{Na2SO4 + 2C -> Na2S + 2CO2}.$$


First I calculated molecular mass of $\ce{Na2S}$ which is $\pu{78.04 g/mol}$.


Then I divided to get the amount of substance: $\frac{58}{78.04} = \pu{0.743 mol}$.


The ratio between $\ce{C}$ and $\ce{Na2S}$ is $2 : 1$, so the amount of substance of $\ce{C}$ is $0.743 \cdot 2 = \pu{1.486 mol}$.


Now I multiplied the amount of substance of $\ce{C}$ with the molar mass of $\ce{2C}$ to get the mass: $1.486 \cdot 24.022 = \pu{35.69 g}$


I submitted this result and got a wrong answer. I don't really understand what I did wrong here. What am I missing? (Assume no limiting factor).




Echoing Martin's excellent comment in answer form, I think it is very helpful to write out these units conversion problems in ["railroad tracks"](https://www.katmarsoftware.com/articles/railroad-track-unit-conversion.htm) format.


The only trick with chemistry and stoichiometry when it comes to unit conversions is that there are often multiple species whose mass or moles (or volume or density etc) we are tracking. You'd better write down species labels in your units. And when you do this, a `mol of Na2S` is different than a `mol of reaction`; the two mole units do NOT cancel out if you take the ratio of one to the other.


The stoichiometry of the particular reaction you're interested in is what lets you convert from moles of one species to another. So your particular reaction is telling you that `1 mol of Na2S` is equal to `1 mol of reaction`. But if you had a different reaction, the "units conversion" between moles of one species and moles of another could be different.


For this particular problem, I'd use the "railroad tracks" method to write out the solution like this (n.b. this is hard to do in MathJax; apologies if I do anything wrong):


1. You start off with 58 g of $\ce{Na2S}$ and the idea to convert it to moles is a good one.


$$\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}}$$


2. Once you have moles of $\ce{Na2S}$, you can convert to moles of "reaction" using the stoichiometric coefficient of $\ce{Na2S}$ in your reaction, which in this case for $\ce{Na2SO4 + 2C -> Na2S + 2CO2}$ is 1.
$$=\require{cancel}\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}$$


And once things are on a "moles of reaction" basis, you can easily "convert" units to the species that you care about, in this case, carbon.


$$=\require{cancel}\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}\frac{\pu{2mol C}}{\pu{mol reaction}}$$


3. Lastly, you can convert from moles of carbon to grams.


$$=\frac{\pu{58 g \ce{Na2S}}}{1}\frac{\pu{mol \ce{Na2S}}}{\pu{78.08 g \ce{Na2S}}} \frac{\pu{mol reaction}}{\pu{mol \ce{Na2S}}}\frac{\pu{2mol \ce{C}}}{\pu{mol reaction}}\frac{\pu{12 g \ce{C}}}{\pu{mol \ce{C}}}$$


4. The final step is to check that all units cancel except the output units you desire, in this case grams of carbon.


$$\require{cancel}=\frac{58 \cancel{\pu{g \ce{Na2S}}}}{1} 
 \frac{\cancel{\pu{mol \ce{Na2S}}}}{78.08\cancel{\pu{ g \ce{Na2S}}}} 
 \frac{\cancel{\pu{mol reaction}}}{\cancel{\pu{mol \ce{Na2S}}}} 
 \frac{2 \cancel{\pu{mol \ce{C}}}}{\cancel{\pu{mol reaction}}}. 
 \frac{\pu{12 g \ce{C}}}{\cancel{\pu{mol \ce{C}}}}$$


$$=\left(\frac{58}{78.04}\frac{2}{1}\frac{12}{1}\right) \textrm{g}\;\ce{C} =\textrm{17.84 g C}$$


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