Consider the reaction: 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) a. If 285.5 mL of SO 2 reacts with 158.9 mL of O 2 (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and the theoretical yield of SO 3 ?

Hey everyone, Let's check out this problem at STP, calculate the mass of phosphene that can be formed from the reaction of 42.5 g of phosphorus and 96 liters of hydrogen gas. Now, STP stands for standard temperature and pressure in our standard temperature is equal to 273.15 Kelvin and our standard pressure Is equal to 1 80 M. So the first thing we can do here is balance our equation here. It's not balanced on the left side. We have four phosphorus, so we'll add a four here. And that changes our hydrogen amount on the right side to 12 hydrogen. So we'll add a six here. Now we're going to calculate the amount of product produced from each of the react mints and we'll get to numbers and the lesser amount of product formed is going to be our limiting reactant. And that's going to be the answer here. So let's start off with one of the reactant that we know And that is 42.5 g phosphorus. So we want to go from grams of phosphorus two g of product and we can do that using our balanced equation and our molecular weights. So in one mole Of phosphorus, using our periodic table, we see that phosphorus weighs 31 g and we have four. So that gives us a molecular weight of 124 g and our units will cancel here and we can go from moles of phosphorus two moles of phosphorus in using our multiple ratio in our equation. So in one mole of phosphorus we have four moles of phosphorus Our units cancel. And we want to go from most of phosphorus 2g of phosphorus and we'll do that using the molecular weight of phosphorus. So in one mole of phosphene we have Phosphene is and we have hydrogen and We have three of them in each weighs one. So we get a total of 34 g. Alright our units cancel and we can go ahead and do our math here and when we do that, We get 46 . g of phosphorus. So that's for one of our reactant. We need to do the same thing but we'll do it for our hydrogen gas and for hydrogen gas they tell us we have 96 liters. And so we can use our ideal gas equation to solve for this one. So are ideal gas equation is PV equals N. R. T. And we want to solve for N. Which will give us moles of hydrogen gas and from moles of hydrogen gas we can go to grams of phosphorus in. So let's go ahead and start off by dividing both sides and isolating our N. So we get and is equal to PV over R. T. And we can go ahead and plug in what we know, We know that our pressure is 180 M. Our volume is given 96 leaders, R r is r gas constant 0.08206 leaders atmosphere over more Calvin. And our temperature Is our standard temperature of 273.15 Kelvin. So let's make sure our units cancel. Are ATMs cancel, our leaders cancel and our kelvin's cancel. And we're left with moles. Let me make this a little smaller. Okay, so once we do this we get n is equal to 4. Moles of H two. Now we need to go from moles of H22g of our product phosphene. So we'll do the same thing using dimensional analysis in our balanced equation. So Um are multiple ratio between our hydrogen gas and phosgene is 4-6 in six moles of hydrogen gas. Get this from our balanced equation. We have four moles of phosphene And in one mole of phosgene our molecular weight Is 34 g. Okay, so make sure our units cancel moles of hydrogen gas cancel and moles of phosphate in cancel. And we're left with a weight of 97 grams of phosphorus. So now we can see how much each of our reactant sites produce. And the one that is the lesser amount is called are limiting reactant. So here we have 46.6 g of phosphene is produced from our phosphorus and this is going to be our correct answer. And that is the end of this problem. I hope this was helpful

Ch.6 - Gases

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Problem 104a

Chapter 6, Problem 104a

Consider the reaction:
2 SO₂(g) + O₂(g) → 2 SO₃(g)
a. If 285.5 mL of SO₂ reacts with 158.9 mL of O₂ (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and the theoretical yield of SO₃?

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Convert the volumes of SO2 and O2 to moles using the ideal gas law: PV = nRT. Use the given conditions (P = 50.0 mmHg, V = 285.5 mL for SO2 and 158.9 mL for O2, T = 315 K) and R = 62.36 L mmHg/mol K.

Calculate the moles of SO2 using the ideal gas law: n(SO2) = (P * V) / (R * T).

Calculate the moles of O2 using the ideal gas law: n(O2) = (P * V) / (R * T).

Determine the limiting reactant by comparing the mole ratio from the balanced equation (2 SO2 : 1 O2) with the calculated moles of SO2 and O2.

Calculate the theoretical yield of SO3 using the stoichiometry of the balanced equation and the moles of the limiting reactant.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limiting Reactant

The limiting reactant is the substance that is completely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. To identify the limiting reactant, one must compare the mole ratios of the reactants based on the balanced chemical equation. In this case, the reaction of sulfur dioxide (SO2) and oxygen (O2) to form sulfur trioxide (SO3) requires a specific stoichiometric ratio.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions based on the balanced equation. It involves using mole ratios derived from the coefficients in the balanced equation to determine how much of each reactant is needed or how much product can be formed. In this scenario, stoichiometry will help in calculating the theoretical yield of SO3 from the given volumes of SO2 and O2.

Gas Laws

Gas laws describe the behavior of gases in relation to pressure, volume, temperature, and the number of moles. The ideal gas law (PV=nRT) is particularly useful for converting the given volumes of gases at specific conditions into moles, which are necessary for stoichiometric calculations. Understanding how to apply gas laws is essential for determining the amounts of reactants and products in reactions involving gases.

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Combined Gas Law

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Consider the reaction: 2 SO2(g) + O2(g) → 2 SO3(g) a. If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and the theoretical yield of SO3?

Verified Solution

Key Concepts

Limiting Reactant

Stoichiometry

Gas Laws

Consider the reaction:
2 SO₂(g) + O₂(g) → 2 SO₃(g)
a. If 285.5 mL of SO₂ reacts with 158.9 mL of O₂ (both measured at 315 K and 50.0 mmHg), what is the limiting reactant and the theoretical yield of SO₃?