For each of the following unbalanced chemical equations, suppose 25.0 g of each reactant is take

For each of the following unbalanced chemical equations,...

For each of the following unbalanced chemical equations, suppose $25.0 \mathrm{g}$ of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface. a. $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ b. $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathbf{N} \mathbf{O}(g)$ c. $\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathbf{N a C l}(a q)$ d. $\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathbf{N} \mathbf{H}_{3}(g)$

For each of the following unbalanced chemical equations, suppose $25.0 \mathrm{g}$ of each reactant is taken. Show by calculation which reactant is limiting. Calculate the theoretical yield in grams of the product in boldface.
a. $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathbf{C} \mathbf{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$
b. $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathbf{N} \mathbf{O}(g)$
c. $\mathrm{NaClO}_{2}(a q)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{ClO}_{2}(g)+\mathbf{N a C l}(a q)$
d. $\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightarrow \mathbf{N} \mathbf{H}_{3}(g)$

Key Concepts

Limiting Reactant

This concept refers to the substance in a chemical reaction that is completely consumed first, thus determining the maximum amount of product that can be formed. In stoichiometric calculations, identifying the limiting reactant is essential because it limits the extent of reaction and the overall yield, regardless of the presence of other reactants.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be formed from a given amount of limiting reactant, as predicted by the balanced chemical equation. It is calculated by converting the quantity of the limiting reactant to moles, using the stoichiometric ratios to find the moles of product, and then converting these moles back to grams using the product's molar mass.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction, as dictated by the balanced chemical equation. It provides a systematic method to calculate amounts of reactants required or products formed, typically involving conversion between grams, moles, and molecules.

Mole-to-Gram Conversions

This concept involves converting between the mass of a substance and the number of moles using its molar mass. Such conversions are crucial in stoichiometry, as they allow calculation of the amount of a substance in moles which can then be related to other substances in the reaction based on the balanced equation.

Balancing Chemical Equations

A balanced chemical equation shows the conservation of mass by ensuring that the number of atoms for each element is equal on both sides of the reaction. Balancing is a critical step in stoichiometric calculations because the coefficients in the balanced equation provide the ratios needed to determine the limiting reactant and theoretical yield.

Transcript

00:01 In order to calculate the mass of the product in bold, we need to first identify the limiting reactant.

00:08 And before we can identify the limiting reactant, we need to have a balanced chemical reaction.

00:16 The first chemical reaction is c2h508 reacting with oxygen, producing carbon dioxide and water in this combustion reaction.

00:24 We see that we have two carbons and one carbon, so we need to put a two in front of the carbon dioxide.

00:30 We have six hydrogens.

00:32 We have two hydrogens.

00:34 So we need to put a three in front of the water.

00:37 Then we'll look at how many oxygens we have.

00:40 We have three and four.

00:44 So that's seven oxygens.

00:46 And we have three over here.

00:48 So if we put a three right there, that's going to give us six plus one, seven.

00:54 And now it's balanced.

00:55 With a balanced chemical reaction and 25 grams of each reactant, we can calculate the amount of product in bold that can be created from 25 grams of each of them.

01:08 The one that produces the least amount will be the limiting reactant, and the amount of that product created will come from the limiting reactant.

01:16 So if we have 25 grams of c2h -50h, we can convert it into moles by dividing by its molar mass, then we'll recognize the stoichiometry is 1 mole of it for every 2 moles of 1 .5 .0 .4.

01:30 Carbon dioxide, the species in bold.

01:32 Then when we have moles carbon dioxide, we can convert to grams carbon dioxide by multiplying by the molar mass carbon dioxide, and we get 47 .8 grams carbon dioxide.

01:43 Let's now look at the 25 grams of 02, divide by its molar mass to get moles oxygen, then recognize three moles ' oxygen produces two moles carbon dioxide.

01:53 Then again, when we have moles carbon dioxide, we multiply by the molar mass carbon dioxide to get the grams carbon dioxide.

02:01 22 .9 grams carbon dioxide.

02:04 So because oxygen produced the least amount of carbon dioxide, it is the limiting reactant, and the amount of carbon dioxide created comes from it.

02:16 Then for part b, the reaction is nitrogen gas reacting with oxygen gas, producing nitrogen monoxide.

02:22 We have two nitrogens, two oxygens, so if we put a two there, then we're balanced.

02:29 We'll start with the 25 grams of nitrogen, converted into moles nitrogen, by dividing by the molar mass nitrogen, then recognize the stoichiometry is 1 to 2, and we'll get moles of nitrogen, nitrogen monoxide.

02:45 Then when we have moles nitrogen monoxide, we'll multiply by the molar mass nitrogen monoxide to get grams nitrogen monoxide, and we get 53 .6 grams.

02:55 But we also need to figure out how many grams nitrogen monoxide can be created from 25 grams of oxygen, divide by the molar mass of oxygen, multiply by 2 because we produce 2 moles nitrogen monoxide for every mole of oxygen.

03:12 Then when we have moles nitrogen monoxide, we'll multiply by its molar mass, as we did up here, to get the grams that would be created if all 25 grams of oxygen were consumed.

03:22 And we get 46 .9 grams nitrogen monoxide.

03:26 So because oxygen produces the least amount of nitrogen monoxide, it is the limiting reactant, and the mass of nitrogen monoxide comes from it, 46.

03:35 0 .9 grams.

03:37 Next we have sodium chloride reacting with chlorine to produce chlorine dioxide and sodium chloride.

03:48 I'm going to put a two here although we have two chlorines and two chlorines and we had one sodium and one sodium this is going to be helpful.

04:01 Now that i put the two there i'm going to put a two here which will give me two chlorines here and two chlorines here, but i still have another two chlorines there...

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