I want to react 250 g of $\ce{Na}$ and enough $\ce{Cl2}$ to produce $\ce{NaCl}$ and see how many grams of $\ce{NaCl}$ I can get.

I guess I should create a scheme like this

               2 Na   +   1 Cl_2   -->   2 NaCl
Number Ratio   2          1              2
Molar Mass    23.0       71.0           58.5
Mass Ratio     250        ?              ?

But I don't know what to put in the two remaining cells.

I think it should be $\frac{250}{2 \cdot 23/71} = 385.5$ and $\frac{250}{23/58.5} = 635.5$, but I am not entirely sure though.

When should I be careful about limiting reagents?


               2 Na   +    1 Cl_2   -->   2 NaCl
Number Ratio   2           1              2
Molar Mass    23.0        71.0           58.5
Moles          250/23=10.9 ?              ?          
Mass Ratio     250         ?              ?

Is the number of moles for the two other reactants just $10.9 / 2 = 5.5$ and $10.9$, respectively?

If so I can easily find the masses with $n = m/M \rightarrow m = n M$.

Edit 2

Now I understand it :-)

But in what situations should I be careful about limiting reagents?

  • $\begingroup$ Add an extra row to your table and call it moles. Work out the number of moles of sodium present. You need half the number of moles of chlorine molecules (remembering that there are two chlorine atoms in a molecule). Since you know the molecular weight of the chlorine molecule you can fill in your gap and similarly for sodium chloride. $\endgroup$ – user1945827 Oct 13 '16 at 10:06
  • $\begingroup$ I have updated my question with the row you suggested $\endgroup$ – Jamgreen Oct 13 '16 at 10:14
  • $\begingroup$ So, 10.9 represents the amount of sodium. Can you work out how many moles of chlorine molecules and sodium chloride are present? $\endgroup$ – user1945827 Oct 13 '16 at 10:24

A typical problem on the concept of limiting reagents!

I'm not sure about you, but I'd go about finding the answer this way; step-by-step.

A) List out the (gram) molar masses of both your reactants.

$Na$ = 23 and $\mathrm{Cl_{2}}$ = 71

B) Now gauge the proportions of reactant in your final product.

The final product is NaCl. That's Na to Cl in the ratio of 1:1 (moles). Taking the masses into consideration, that's 23 : 35.5 grams, or if you want, you can further simplify it by dividing both terms by 23. So the resultant ratio is 1 : 1.54.

Essentially, what this ratio tells you is that, while synthesizing NaCl, 1.54 grams of Cl combines with every gram of Na.

You're given a specific amount of Na, and no constraint has been set on the amount of $\mathrm{Cl_{2}}$ you can use. So here, Na is the limiting reagent.

As you have 250 grams of Na, the corresponding amount of Cl you need is about (1.54)x(250) = 385 grams.

But since you're dealing with chlorine gas ( $\mathrm{Cl_{2}}$ ), this corresponds to 385/71 = 5.42 moles of $\mathrm{Cl_{2}}$ .

C) Take the ratio of molar masses between the limiting reagent and the final product.

Now the gram molar mass of NaCl is 58.5 grams. Now if you take the ratio between the gram molar masses of Na and NaCl, what you get is 23 : 58.5. Simplifying it by dividing both terms by 23, we get 1: 2.54. So for every gram of Na you have that completely react with chlorine, 2.54 grams of NaCl is produced.

Since you have 250 grams of Na, if you completely react it with Chlorine gas (about 385 grams of it), you get (2.54)x(250) = 635 grams of NaCl. Or alternatively, since you now have the mass of chlorine, using the law of conservation of mass you can go on right ahead and simply add 250g Na + 385g $\mathrm{Cl_{2}}$ and get 635g NaCl. Voila!

[Yes I know I made it longer than it could've been, but hopefully going step by step won't create any confusion. You don't really have to calculate the amount of chlorine as well, but I did it anyway.]

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