Consider the combustion reaction of 0.148 g of a hydrocarbon having formula Cn H2 n+2 with an ex

Consider the combustion reaction of 0.148 g of a hydrocarbon...

Consider the combustion reaction of 0.148 $\mathrm{g}$ of a hydrocarbon having formula $\mathrm{C}_{n} \mathrm{H}_{2 n+2}$ with an excess of $\mathrm{O}_{2}$ in a 400.0 $\mathrm{mL}$ steel container. Before reaction, the gaseous mixture had a temperature of $25.0^{\circ} \mathrm{C}$ and a pressure of 2.000 atm. After complete combustion and loss of considerable heat, the mixture of products and excess $\mathrm{O}_{2}$ had a temperature of $125.0^{\circ}$ C and a pressure of 2.983 atm. (a) What is the formula and molar mass of the hydrocarbon? (b) What are the partial pressures in atmospheres of the reactants? (c) What are the partial pressures in atmospheres of the products and the excess $\mathrm{O}_{2} ?$

Consider the combustion reaction of 0.148 $\mathrm{g}$ of a hydrocarbon having formula $\mathrm{C}_{n} \mathrm{H}_{2 n+2}$ with an excess of $\mathrm{O}_{2}$ in a 400.0 $\mathrm{mL}$ steel container. Before reaction, the gaseous mixture had a temperature of $25.0^{\circ} \mathrm{C}$ and a pressure of 2.000 atm. After complete combustion and loss of considerable heat, the mixture of products and excess $\mathrm{O}_{2}$ had a temperature of $125.0^{\circ}$ C and a pressure of 2.983 atm.
(a) What is the formula and molar mass of the hydrocarbon?
(b) What are the partial pressures in atmospheres of the reactants?
(c) What are the partial pressures in atmospheres of the products and the excess $\mathrm{O}_{2} ?$

Key Concepts

Dalton’s Law of Partial Pressures

Dalton’s Law states that in a mixture of non-reacting gases, the total pressure is the sum of the partial pressures of the individual gases. This concept is key when analyzing the gaseous products and excess oxygen in a combustion reaction, as it allows the separation of the contributions to the total pressure from each component gas.

Ideal Gas Law

The Ideal Gas Law (PV = nRT) is a fundamental principle that links the pressure, volume, temperature, and number of moles of a gas. It is crucial in problems where gaseous reactants and products are involved, allowing one to calculate changes in pressure related to changes in temperature and the number of moles, as well as to determine partial pressures in a closed system.

Combustion Reactions

This concept involves understanding how organic compounds, specifically hydrocarbons, react with oxygen to produce carbon dioxide and water. It requires the ability to write and balance the chemical equation, which is essential for determining the stoichiometric relationships between reactants and products.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. In combustion analysis, it is used to relate the mass of the reactant (the hydrocarbon in this case) to the number of moles of oxygen consumed and the moles of products formed, thereby helping in determining molecular formulas and molar masses.

Transcript

00:01 This is a very important problem in your textbook because not only does it allow us to practice the behavior of gases and introduce us to the chemistry of gases, but this allows us to be introduced to a very important concept in chemistry, which is if we have a reaction, what's going to happen if we have an unreacted product or reactive or excess of one of those compounds? and that's what we're going to be investigating in this problem.

00:33 Let's first note that the pressure of the ieth gas is given by this equation.

00:38 P sub i is equivalent to n sub i divided by n total times p t.

00:44 All that this equation is saying is that p sub i, the pressure of some gas, is equivalent to the number of moles of that gas, divided by the total number of moles times the total pressure.

00:57 However, at if constant t and v, temperature and pressure, or pardon me, temperature and volume, p over n is equal to a constant.

01:08 So what that is saying to us is that pressure is directly related or directly proportional to n, or number of moles.

01:19 Now, this is a very important problem because it introduces us to what's called an ice table, where we can control or organize our initial and final number of moles of our reaction.

01:32 So we're given this reaction of cs2 plus 302 that form co2 and two moles of so2.

01:41 I stands for our initial number of moles.

01:45 F is our final number of moles.

01:47 So for cs2, we start with a moles and we are left with zero.

01:52 For o2, we start with 3a plus b.

01:56 We are left with b.

01:59 For co2, we start with zero because remember this is a reactant, or pardon me, this is a product.

02:05 We will gain a.

02:07 For s .o2, we'll start with zero, and we will gain 2a...