Consider the reaction for the production of NO2 from NO : 2 NO(g)+O2(g) ⟶2 NO2(g) (a) If 84.8 L

step 1 In this question, we have NO and O2 reacting to make NO2. And in Part A, we have 84 .8 liters of

Consider the reaction for the production of NO2 from NO : 2...

Consider the reaction for the production of $\mathrm{NO}_{2}$ from $\mathrm{NO}$ : $$ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) $$ (a) If $84.8 \mathrm{L}$ of $\mathrm{O}_{2}(g),$ measured at $35^{\circ} \mathrm{C}$ and $632 \mathrm{mm} \mathrm{Hg}$ is allowed to react with $158.2 \mathrm{g}$ of $\mathrm{NO}$, find the limiting reagent. (b) If 97.3 L of $\mathrm{NO}_{2}$ forms, measured at $35^{\circ} \mathrm{C}$ and $632 \mathrm{mm}$ Hg, what is the percent yield?

Consider the reaction for the production of $\mathrm{NO}_{2}$ from $\mathrm{NO}$ :
$$
2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)
$$
(a) If $84.8 \mathrm{L}$ of $\mathrm{O}_{2}(g),$ measured at $35^{\circ} \mathrm{C}$ and $632 \mathrm{mm} \mathrm{Hg}$
is allowed to react with $158.2 \mathrm{g}$ of $\mathrm{NO}$, find the limiting reagent.
(b) If 97.3 L of $\mathrm{NO}_{2}$ forms, measured at $35^{\circ} \mathrm{C}$ and $632 \mathrm{mm}$ Hg, what is the percent yield?

Key Concepts

Limiting Reagent

The limiting reagent is the reactant that is completely consumed in a chemical reaction, thereby limiting the amount of product produced. Determining the limiting reagent involves calculating the moles of each reactant and using the molar ratios from the balanced equation to see which one will run out first.

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate the moles of reactants to the moles of products. It allows you to predict the amount of product that can be formed from a given amount of reactants, ensuring the conservation of mass and moles in a chemical reaction.

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure, volume, temperature, and number of moles of a gas. When working with gases under non-standard conditions, this law is used to convert measured volumes to moles, which can then be used in stoichiometric calculations.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction and is calculated by dividing the actual yield by the theoretical yield (the maximum amount of product possible from the limiting reagent) and then multiplying by 100. This comparison highlights any loss of material or incomplete reaction.

Transcript

00:05 In this question, we have no and o2 reacting to make no2.

00:11 And in part a, we have 84 .8 liters of oxygen at 35 degrees celsius and 632 millimeters of mercury, reacting with 158 .2 grams of no, and we're asked to find the limiting reagent.

00:25 So what we're going to do is use the information given to us to find the number of moles of no2 that would be produced from this quantity of oxygen and this.

00:35 Quantity of no2 or of no and whichever one gives us the lower amount of no2, that will be the limiting reagent since that will be used up first.

00:44 So for the o2, i am going to use the ideal gas equation to find the number of moles of this.

00:51 So that is pv equals nrt, where p is pressure, v is volume, n is the number of moles, r is a constant, and t is temperature.

01:02 But first i need to make that everything is in the correct units.

01:06 So my temperature needs to be in kelvin.

01:08 So to get from celsius to kelvin, i add 273 and i get 308 kelvin as my temperature.

01:16 And my pressure needs to be in atmospheres.

01:17 So to get from millimeters of mercury to atmospheres, i'm going to divide by 760, and i will get 0 .832 atmospheres as my pressure.

01:27 And then this is my volume.

01:29 It's already in liters, so we're good there.

01:31 So i'm just going to plug this into the ideal gas equation.

01:33 And solve for n.

01:35 So my pressure is 0 .832 atmospheres.

01:41 My volume is 84 .8 liters.

01:46 We're looking for n.

01:47 R is 0 .0821 liter atmosphere per mole kelvin and my temperature is 308 kelvin.

01:57 So if i solve this for n, i get an n of 2 .79 moles.

02:08 Of 02 and then i'm going to convert that to moles of n02.

02:13 So for two moles of n02 i will have one mole of o2 according to my balanced reaction of above.

02:24 So moles of o2 cancels out and that gives me 5 .58 moles of n02 that would be produced here.

02:34 And then for the n0 we're going to use blue.

02:38 We're going to do this one in blue.

02:40 Blue and we're just going to go from grams of this to moles of it to moles of it...

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