Limiting Reactant

Question

If 11.54 grams of pure sodium is reacted in the presence of 21.4grams of chlorine gas

 The limiting reactant in this situation is....

1. a. NaCl
2. b. Na
3. c. Cannot tell from information given
4. d. Cl₂

Answer 1

Hello, Anslie,

We are asked whether we can determine if there is a limiting reagent under the circumstances noted. We are given masses, but those don't tell us much, yet. If we convert them into moles, we're better able to compare the actual numbers of molecules in each sample. Let's do that now:

Moles Na: (11.54 grams Na)/(23.0 grams/mole Na) = 0.502 moles Na

Moles Cl₂: (21.4 grams Cl₂)/(70.9 grams/mole Cl₂) = 0.302 moles Cl₂

We have many more moles of Na than Cl₂, but to make a correct comparison, we need a balanced eqaution to find the molar ratio of Na/Cl₂ in a complete reaction.

I get:

2Na + 1Cl₂ = 2NaCl

Now we're getting close. The balanced equation tells us we need twice as many moles of Na as we do of Cl₂. We can see just by looking that we won't have 2*(0.302 moles Cl₂) = 0.604 moles Na. We only have 0.501 moles of sodium.

Sodium is therefore the limiting reagent. Once it is gone (0.501 mole) it will have consumed (1/2)(0.502)=0.251 moles of the chlorine. That will leave (0.302 - 0.251=) 0.051 moles of chlorine unreacted.

NaCl is a product, so cannot be a limiting reagent. It is simply limited to the amount of Na present. We might also note that it has a coefficient of 2 in the balanced equation. Since Na is the limiting reagent, and we produce 2 moles of salt for every 2 moles of sodium (a molar ratio of 1:1), we should produce 0.502 moles of NaCl. That's 29.3 grams of NaCl.

Bob