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Limiting reagents question

What mass of calcium hydroxide is formed when 10.0g of calcium oxide reacts with 10.0g of water?
CaO + H2O - - > Ca(OH)2

ik this should be simple but my brain isn't working right now. Even if u give an example instead it would be helpful. :smile:

Reply 1

Yuno2001

Original post by michaelahahah

Hi Michael, I hope you're having a great day! :smile:

To find the limiting reagent, you'll have to first calculate the moles of each reactant, in this case, the moles of calcium oxide and the moles of water molecules provided.

n(CaO)= 10/ (Mr(CaO))
n(H2O)= 10/ (Mr(H2O))

The equation tells you the ratio at which the two compounds combine, which is 1:1 in this case.
In this question, the reactant present in the least amount is the limiting reactant.

In this case, CaO is the limiting reactant, as It will limit the number of water molecules that will be used up in the reaction. You should also note that there will be leftover (unused) water molecules after the reaction.

n(CaO) = n(H2O) used = n(Ca(OH2)) formed

n(H2O) leftover = n(H2O) provided - n(H2O) used

In other cases, however, you may have to do a further calculation to find the limiting reactant.
Take a look at this reaction.

Say, find the limiting reagent of this reaction, given 20g of Mg reacts with 36g of HCl.

Some ppl may think that magnesium is the limiting reactant straight away, thinking, "The reactant present in the least amount is the limiting reagent."

But, look at the ratio at which the magnesium and hydrochloric acid combine. 1:2 right?
To find the limiting reagent, divide the moles of each reactant by the coefficient of the balanced equation. The lowest number indicates the limiting reagent.

I hope this helps! Have a great one! :smile:

Reply 2

snehalb

Original post by michaelahahah

hey there! i see your question has already been answered very meticulously by a fellow user! but if you're anything like me, an easier to understand answer would help, so I'm writing one :smile:

If you want to use the other one by all means ignore mine, no worries :smile:

First we would calculate the moles of both CaO and H2O.
Moles = Mass/ Molar Mass

The molar mass of CaO is 56 so number of moles in 10g = 10/56 which is 0.178 moles
The molar mass of H2O is 18 so number of moles in 10g = 10/18 which is 0.555 moles

Observing the equation we know that 1 mole of CaO requires 1 mole of H2O. Thus 0.555 moles of H2O would require 0.555 moles of CaO but oh no! that's not available :frown:

Thus, we realise that CaO is the limiting reagant. It limits the reaction such that only 0.178 moles of H2O are used up.

This means 0.178 moles of Ca(OH)2 are produced! The molar mass of Ca(OH)2 is 74 g and thus the mass would be 74 x 0.178 = 13.172 g

Hope this helped in any way possible!!