May be solved by the molar volume method (Section 14.7 ) or...

May be solved by the molar volume method (Section 14.7 ) or by the ideal gas equation method (Section 14.8 ). In the answer section, the setups are given first for the molar volume method, printed over a tan background. Then, over a green back- ground, the answers are given for the ideal gas equation method. Check your work according to the section you studied. What volume of hydrogen gas at $29^{\circ} \mathrm{C}$ and 0.809 atm is produced when $31.8 \mathrm{g}$ of iron reacts completely according to the reaction $\mathrm{Fe}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{FeCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) ?$

May be solved by the molar volume method (Section 14.7 ) or by the ideal gas equation method (Section 14.8 ). In the answer section, the setups are given first for the molar volume method, printed over a tan background. Then, over a green back-
ground, the answers are given for the ideal gas equation method. Check your work according to the section you studied.
What volume of hydrogen gas at $29^{\circ} \mathrm{C}$ and 0.809 atm is produced when $31.8 \mathrm{g}$ of iron reacts completely according to the reaction $\mathrm{Fe}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{FeCl}_{2}(\mathrm{aq})+\mathrm{H}_{2}(\mathrm{g}) ?$

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate the quantities (in moles) of reactants and products. This concept allows one to determine how much product is formed from a given amount of reactant, ensuring that the law of conservation of mass is obeyed in the reaction process.

Mole Concept

The mole concept is fundamental in chemistry, providing a bridge between the mass of a substance and the number of its constituent particles. It involves converting a given mass of a substance into moles using its molar mass, thereby enabling the application of balanced chemical equations to predict the amounts of products or reactants involved.

Ideal Gas Law

The ideal gas law, expressed as PV = nRT, is a key relationship that connects pressure, volume, temperature, and the number of moles of a gas. This principle is essential for calculating any one of the gas properties when the others are known, and it is widely used in chemical problems that involve gas reactions or gas collection under non-standard conditions.

Molar Volume of a Gas

The molar volume concept states that one mole of any ideal gas occupies a specific volume under given temperature and pressure conditions. This idea simplifies calculations involving gases at standard conditions and serves as an alternative approach to the ideal gas law, making it easier to estimate the volume of a gas produced or consumed in a reaction.

Transcript

00:02 Hi there.

00:03 In this problem, i am going to start out by writing our balanced equation.

00:08 We are given the balanced equation in the problem.

00:11 It is fe, combines with 2hcl to produce fecl plus h2, with the h2, of course, being a gas.

00:35 We are starting with 31 .8 grams of the iron.

00:43 And what we want to determine is how many liters of the hydrogen we're going to produce at 29 degrees celsius and 0 .809 atmospheres of pressure.

01:05 I'm going to go ahead and convert my celsius to kelvin right away by adding 273.

01:11 That will give me 302 kelvin.

01:18 Okay, to plan this problem out.

01:20 It looks like i can do some stoichiometry.

01:23 I can start with grams of iron and convert to of iron and then convert from moles of iron to moles of hydrogen.

01:31 At that point, i cannot use 22 .4 because my temperature and pressure are not at standard temperature pressure.

01:40 So what i can do at that point is take the moles of hydrogen and convert to volume using the ideal gas equation.

01:48 Alright, so let's get started.

01:49 I'm going to do this in two steps.

01:50 So first i'm going to do the stoichiometry part.

01:53 So i'm starting with 31 .8 grams of the iron.

01:56 My first step is to divide by the molar mass of iron.

02:02 In every one mole of iron, there are 55 .85 grams of the iron.

02:18 This will allow me to cancel grams of iron.

02:22 My next step, i'm going to use the mole ratio from the balanced equation.

02:27 The coefficients in front of iron and h2 in this balanced equation are both ones.

02:34 So for every one mole of iron, i'm going to produce 1...

Please give Ace some feedback