Limiting Reactants

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VGonzalez
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Joined: Fri Sep 28, 2018 12:18 am

Limiting Reactants

Postby VGonzalez » Wed Oct 03, 2018 1:22 pm

In the class example involving solid calcium carbide to form ethyne gas Mr. Lavelle wrote that 1.56 moles of calcium carbide requires 3.12 moles of water. What does this mean? Is it just about getting to the closest whole number? Does it have any affect on solving the problem?

Josephine Chan 1B
Posts: 31
Joined: Fri Sep 28, 2018 12:17 am

Re: Limiting Reactants

Postby Josephine Chan 1B » Wed Oct 03, 2018 1:29 pm

since the ratio of solid calcium carbide to water is 1:2(refer to balanced equation), 1.56 mol of calcium carbide requires 2 x 1.56 = 3.12 mol of water which is less than the 5.55 mol of water that we have. That means we have excess water and not enough calcium carbide. So calcium carbide is the limiting reactant.

Jewelyana3A
Posts: 31
Joined: Fri Sep 28, 2018 12:24 am

Re: Limiting Reactants

Postby Jewelyana3A » Wed Oct 03, 2018 1:52 pm

Just like what was said already, it is all about ratios. Always remember to balance the equation first so that you can easily refer to it later on in the problem. So, the first part of the question was identifying the limiting reactant, which is between water or calcium carbide. You have to figure out the moles of both calcium carbide and water. This leads you to 1.56 mol of calcium carbide and 5.55 mol of water. (That is your calculated amount). Now, you must compare it to your required amount by basically applying it to the ratio. Notice that there is less mol of calcium carbide than there is of water. (Also, it helps to remember the illustration mentioned of tires and assembling cars. If you have not enough tires, but an excess of other material then the tires are the "limiting reactant" in a sense.) Due to the 1:2 ratio 1.56 mol of calcium carbide requires 3.12 mol of water. Comparing moles there is too much water (an excess) and not enough calcium carbide. Hopefully this makes sense!


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