In the reaction #"FeC"l_2 + 2"NaOH" -> "Fe(OH)"_2"(s)" + 2"NaC"l#, if 6 moles of #"FeC"l_2# are added to 6 moles of #"NaOH"#, how many moles of #"FeC"l_2# would be used up in the reaction?

Answer 1

#"3 moles FeCl"_2#

The balanced chemical equation indicates the ratio that must constantly exist between the reactants in any chemical reaction.

As for you, you've

#"FeCl"_ (2(aq)) + color(blue)(2)"NaOH"_ ((aq)) -> "Fe"("OH")_ (2(s)) darr + 2"NaCl"_ ((aq))#

The stoichiometric coefficients for sodium hydroxide and iron(II) chloride, respectively, indicate the mole ratio that needs to be present for this reaction to occur.

Notice that you have a #1:color(blue)(2)# mole ratio between the two reactants, so you can say that the reaction will always consume twice as many moles of sodium hydroxide than moles of iron(II) chloride.
Now, you know that #6# moles of iron(II) chloride are added to #6# moles of sodium hydroxide.

Calculate the number of moles of iron(II) chloride that will react with the moles of sodium hydroxide using the previously mentioned mole ratio.

#6 color(red)(cancel(color(black)("moles NaOH"))) * "1 mole FeCl"_2/(color(blue)(2)color(red)(cancel(color(black)("moles NaOH")))) = "3 moles FeCl"_2#
This tells you that in order for all the moles of sodium hydroxide to react, you need #3# moles of iron(II) chloride. The other #3# moles will not take part in the reaction, i.e. they are in excess.

Thus, one could say that

#"3 moles of FeCl"_2 -> "will react"#
#"3 moles of FeCl"_2 -> "will not react"#

It is evident that all of the sodium hydroxide is used up before any moles of iron(II) chloride have an opportunity to participate in the reaction.

This tells you that sodium hydroxide acts as a limiting reagent, i.e. it limits the amount of iron(II) chloride that takes part in the reaction from #6# moles to #3# moles.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

In the balanced chemical equation ( \text{FeCl}_2 + 2\text{NaOH} \rightarrow \text{Fe(OH)}_2(s) + 2\text{NaCl} ), one mole of ( \text{FeCl}_2 ) reacts with two moles of ( \text{NaOH} ). Therefore, if 6 moles of ( \text{FeCl}_2 ) are added, they would react completely with 12 moles of ( \text{NaOH} ). Hence, all 6 moles of ( \text{FeCl}_2 ) would be used up in the reaction.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7