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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).

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  • $\begingroup$ You made a mistake in the last calculation. Stoichiometric coefficients are not included when calculating mass of a species from amount (moles). They're only used when converting amount of product into amount of reactant or vice-versa. In short, your answer divided by 2 is the right answer. $\endgroup$
    – Sam202
    Nov 12, 2022 at 7:52
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    $\begingroup$ You don't multiply by the molar mass of $\ce{2C}$. You simply multiply by the mass of one carbon as you already accounted for the ratio in the previous step. Note that $\ce{2C}$ doesn't imply two carbons bonded to each other. It only represents the stoichiometric ratio in the reaction. $\endgroup$
    – M.L
    Nov 12, 2022 at 7:52
  • $\begingroup$ I was not aware that stoichiometric coefficients are not included when calculating mass of a species from amount. Now I know. Thank you guys for the answer $\endgroup$
    – Adi Wiesel
    Nov 12, 2022 at 8:07
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    $\begingroup$ You should always use symbols and write actual equations with them, before you put in numbers. Never ever skip the units. You'll catch about 99% of the mistakes you make with such approach. $\endgroup$ Nov 12, 2022 at 14:06

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Echoing Martin's excellent comment in answer form, I think it is very helpful to write out these units conversion problems in "railroad tracks" 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.

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

  1. 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.

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

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

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

  1. 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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