How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C₂H₄Cl₂, can be prepared by reaction of 15.4 g of ethylene, C₂H₄, with 3.74 g of Cl₂?

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Hi everyone. This problem reads a segolene is commonly prepared by the reaction of calcium carbide with water to form a sideline and calcium hydroxide, calculate the mass and grams of a sideline that can be prepared by the reaction of 6.4 g of calcium carbide with 10.5 g of water. Okay. So our goal here is to find the following. Alright, so let's start off by writing out our equation. Alright. And our equation is going to be that are calcium carbide reacts with water to produce a seedling. Plus calcium hydroxide. Okay, so looking at this this chemical equation is unbalanced. And so what we're going to need to do is balance it. So let's take a minute to do that. And when we do that we're going to get the following balanced equation. Alright so now we have our balanced equation. We're going to need to now find the limiting re agent. Okay? Remember that are limiting re agent is completely consumed in the reaction and it controls the amount of product formed. So looking at our reactant, we see we have to react ints, we have calcium carbide and we have water. These are two reactant and we need to figure out out of these two which one is the limiting re agent. Okay So what we're going to need to do is we're told how much we have of each, we were told we have 6.4 g of the calcium carbide and we have 10.5 g of the water. So from these two we need to figure out how much a C. D. Line is produced by both of these react ints. Alright so let's start there. So now we're going to start off with our calcium carbide. So we have 6.4 g of calcium carbide. And our goal is we want to go from grams of calcium carbide to moles. No, excuse me. two g of a seedling. Okay so that is what we're going to do first and then we're going to do the same thing. But for the water. So we have 6.4 g of calcium carbide. We want to go from grams of calcium carbide, two moles of calcium carbide. And we do that using molar mass. So in one mole of calcium carbide, The Molar mass is 64. g of calcium carbide. Alright, so our units cancel and we're left with grams. Alright, so now we're in moles of calcium carbide. We want to go from moles of calcium carbide, two moles of a seedling. And we do that using the molar ratio. So looking at our chemical equation, our balanced chemical equation, we need to look at the ratio between our seedling and our calcium carbide. So we see a 1 to 1 molar ratio for every one mole of calcium carbide consumed, one mole of a sideline is produced. So that is our multiple ratio let's go ahead and write that. So for every one mole of calcium carbide consumed. One mole of a sideline is produced. All right, So now our moles of calcium carbide cancel. And we're left with moles of a sideline. But we want grams of a sideline. So we need the molar mass. Alright, so in one mole of a sideline there is looking at our periodic table and calculating the molar mass, there is 26. 26.0373 g of a seedling. Alright, so now our units of moles cancel. And we're left with grams of a seedling, which is what we want. So let's do this calculation. And when we do it, We get 2.6 g of a seedling. Okay, So when we when we have 6.4 g of calcium carbide. This 6.4 g of calcium carbide produces 2.6 g of a seedling. Now we need to go ahead and do the same thing with water. We need to see how much water. So from 10.5 g of water, how much A segolene does this produce? And from there we'll be able to see which one is our limiting reagent. So let's go ahead and do the same thing. Alright, so in the problem we're told we have 10.5 g of water. Okay, so our goal here is to go from grams of water. two g of a seedling. So we're going to do the exact same thing that we did up above. But we're going to have different numbers here. And so let's go ahead and do that and plug in. All right, So we're going to go from grams of water. Two moles of water using molar mass. So one mole of water. The Molar Mass is 18.015 - eight g of water. So are grams of water cancel. And we're left with moles of water. So now we want to go from moles of water, two moles of a seedling. So we need to look at the multiple ratio. So looking at our balanced chemical equation, we see for every one mole of excuse me, for every two moles of water consumed, one mole of a seedling is produced. So that's our multiple ratio 2-1. So let's go ahead and write that for every two moles of water consumed one mole of a seedling is produced. So our moles of water cancel. And we're left with moles of a seedling. Okay? And we need to go from moles of a seedling to grams of a seedling. Okay, so in one mole of a segolene, our molar mass is 26. 73 g of a seedling. Alright, so let's go ahead and do that calculation. And when we do this calculation, we get 7.59 g of a settling. So that means for every 10.5 g of water consumed 7.59 g of a seed Elina has yielded. So Out of these two, which one is the limiting reagent? Which one is going to run out first is completely consumed in the reaction. The one that's going to be completely consumed in the reaction is the calcium carbide. Right? Because we have 2.6 grams. And so we can say that the calcium carbide forms less a seedling. So it's the limiting re agent. And the mass of a seedling formed by the reaction is 2. g. So the questions question asked us to calculate the mass that can be prepared by the reaction. So our mass we're going to highlight here is 2.6 g. All right. And that is it for this problem. I hope this was helpful.

How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C₂H₄Cl₂, can be prepared by reaction of 15.4 g of ethylene, C₂H₄, with 3.74 g of Cl₂?

Verified Solution

Key Concepts

Stoichiometry

Molar Mass

Limiting Reactant

How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C2H4Cl2, can be prepared by reaction of 15.4 g of ethylene, C2H4, with 3.74 g of Cl2?

Verified Solution

Key Concepts

Stoichiometry

Molar Mass

Limiting Reactant

How many grams of the dry-cleaning solvent 1,2-dichloroethane (also called ethylene chloride), C₂H₄Cl₂, can be prepared by reaction of 15.4 g of ethylene, C₂H₄, with 3.74 g of Cl₂?