In the reaction shown below, 64 g CaC2 is reacted with 64 g H2 O . CaC2(s)+2 H2 O(l) →C2 H2(g)+C

step 1 So this is a limiting reagent problem. And we're given the balanced equation, which is super nic

In the reaction shown below, 64 g CaC2 is reacted with 64 g...

Key Concepts

Theoretical Yield

The theoretical yield is the calculated maximum amount of product that could be formed from given quantities of reactants, as dictated by the stoichiometry of the balanced chemical equation. It assumes complete conversion of the limiting reactant into the product and serves as a benchmark for comparing the efficiency of practical reactions.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thereby limiting the amount of product that can be formed. Identifying the limiting reactant is essential for calculating the theoretical yield because it determines the maximum possible amount of product generated.

Excess Reactant

An excess reactant is a substance present in a quantity greater than what is necessary to completely react with the limiting reactant. Since it is not entirely consumed during the reaction, it does not affect the calculation of the theoretical yield, and any remaining excess can be recovered after the reaction.

Stoichiometry

Stoichiometry involves quantifying the relationships between reactants and products in a chemical reaction using a balanced chemical equation. It helps predict how much product will form from given amounts of reactants by using mole ratios, ensuring that the law of conservation of mass is obeyed.

Balanced Chemical Equation

A balanced chemical equation represents the reaction with equal numbers of atoms for each element on both sides of the equation. This balance is crucial because it underpins stoichiometric calculations, ensuring that the proportions of reactants and products conform to the conservation of mass.

Transcript

00:01 So this is a limiting reagent problem.

00:03 And we're given the balanced equation, which is super nice.

00:07 So what we're going to do is start with part a, which tells us that we're given 64 grams of both starting materials.

00:18 That's a different number of moles for each starting material.

00:21 And i think that's the take -home message here is that mass and substance are not the same thing.

00:26 It depends on the compound.

00:27 How many moles correspond to that mass that's given.

00:31 So the first part is asking us which is the excess reagent and which is the limiting reagent.

00:39 So this is actually going to give us, we can do parts a and b together, actually, and i'll show you why.

00:48 So the way that i do this, when you're given these types of problems, these types of problems are actually very common.

00:53 So you're given amounts of both starting materials and you have to figure out what the limiting reagent.

00:58 Is, right? so it doesn't matter which product you pick to calculate.

01:03 So basically the way that i approach these problems is you calculate, you figure out the theoretical yield of both compounds.

01:12 Right.

01:12 So how many, so let's say how many moles of ethylene gas can you make with 64 grams of calcium carbide? that's going to be one equation that we set up.

01:22 And then how many moles of, or how many grams of ethylene gas can we make? from 64 grams of water.

01:32 So you're going to do two sets of unit conversions.

01:35 And then whichever one gives you the lowest number means that that one's going to run out before the others.

01:43 Right.

01:45 So what's going to happen here is let's start with calcium carbide.

01:50 So we're going to start with 64 grams of calcium carbide.

01:55 And i calculated the molecular weight of that to be, ironically.

02:04 64 .1 grams per mole.

02:13 And i'm not writing this right.

02:15 There we go.

02:16 So one mole of calcium carbide is equal to 64 .1 grams of calcium carbide.

02:27 And then we're going to use the balanced equation to go to, i'm going to pick ethylene gas as the product.

02:33 Part c is going to make us go to calcium hydroxide.

02:35 So whichever one you pick, you're actually answering two parts at once.

02:38 It doesn't of which product you pick.

02:41 So i'm going to use the molar ratio, right? so it's one mole of calcium carbide.

02:46 Nope.

02:47 Again, i just did the same thing.

02:48 One mole of ethylene gas, c2h2.

02:52 That's actually acetylene.

02:53 Why am i saying ethylene? sorry, some of these hydrocarbons that are really small have very similar molecular formulas.

02:59 And it's getting me confused.

03:01 So that's acetylene gas.

03:03 It's c2h2.

03:05 So, right.

03:06 So one mole of calcium carbide.

03:10 So now we're in moles of c2h2 and we need to go to, you i mean, you could actually keep your answer in moles if you want to.

03:19 So i didn't do that because it doesn't tell us what unit to put the theoretical yield in.

03:27 So i actually convert it to grams, but i guess you can leave it in moles if you want to...

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