Limiting Reactant

Moderators: Chem_Mod, Chem_Admin

Posts: 99
Joined: Fri Sep 28, 2018 12:17 am

Limiting Reactant

Postby AlyssaBei_1F » Mon Oct 01, 2018 4:20 pm

On the limiting reactant video, the example is calcium carbonate reacts with water to form calcium hydroxide and ethyne. I understand up to calculating the moles of CaC2 and H2O, but am confused why we have to multiply 1.56 (moles of CaC2) by 2. Can someone help me with this?

Hai-Lin Yeh 1J
Posts: 89
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

Re: Limiting Reactant

Postby Hai-Lin Yeh 1J » Mon Oct 01, 2018 4:32 pm

The reason he says that you multiply 1.56 (moles of CaC2) by 2 is because in the balanced chemical equation, it says that 1 mole of CaC2 reacts with 2 mole of H2O.

1CaC2 (s) + 2H2O (l) -> Ca(OH)2 (aq) + C2H2 (g)

By looking at the stoichiometric coefficients in the chemical equation, we see that for every 1 CaC2, 2 H2O is required. Thus, since you currently have 1.56 moles of CaC2, that means you need 3.12 mol of H2O.

Parth Mungra
Posts: 72
Joined: Fri Sep 28, 2018 12:28 am

Re: Limiting Reactant

Postby Parth Mungra » Mon Oct 01, 2018 9:13 pm

In the reaction it takes 2 moles of water for every 1 mols of calcium carbide. The problem was given with information that will let you get 1.56 mol of CaC2 and 5.55 mol of water. Because of the 1:2 ratio, we have to check to see how many mols of water molecules will be used with the given amount of CaC2. So, multiply 1.56 with the 2 mol H2O to get 3.12 mol H2O. With the information that you calculated earlier, we can find that we do have a sufficient amount of water molecules for this reaction to take place, but because we only have 1.56 mol of calcium carbide, we can only us 3.12 mol of water, making calcium carbide the limiting reactant.

Hope this helps

Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest