Mg metal reacts with HCl to produce hydrogen gas. Mg(s)+2 HCl(a q) ⟶H2(g)+MgCl2(a q) a. What vol

step 1 Okay, so in this question we have two parts, part A and B. Let's start with part A first. So let

Mg metal reacts with HCl to produce hydrogen gas. Mg(s)+2...

Mg metal reacts with HCl to produce hydrogen gas. $$ \mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{MgCl}_{2}(a q) $$ a. What volume, in liters, of hydrogen at $0{ }^{\circ} \mathrm{C}$ and $1.00 \mathrm{~atm}$ (STP) is released when $8.25 \mathrm{~g}$ of $\mathrm{Mg}$ reacts? b. How many grams of magnesium are needed to prepare 5.00 L of $\mathrm{H}_{2}$ at $735 \mathrm{mmHg}$ and $18^{\circ} \mathrm{C}$ ?

Mg metal reacts with HCl to produce hydrogen gas.
$$
\mathrm{Mg}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{MgCl}_{2}(a q)
$$
a. What volume, in liters, of hydrogen at $0{ }^{\circ} \mathrm{C}$ and $1.00 \mathrm{~atm}$ (STP) is released when $8.25 \mathrm{~g}$ of $\mathrm{Mg}$ reacts?
b. How many grams of magnesium are needed to prepare 5.00 L of $\mathrm{H}_{2}$ at $735 \mathrm{mmHg}$ and $18^{\circ} \mathrm{C}$ ?

Key Concepts

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It requires using the balanced chemical equation to determine the mole ratios of reactants to products, which then allows one to calculate the amount of product formed from a given amount of reactant or vice versa.

Molar Mass and Mole Conversions

Molar mass is the mass of one mole of a substance and is used to convert between grams and moles. Understanding how to compute the number of moles from a given mass is crucial in stoichiometric calculations, as it connects the mass of reactants or products to the mole ratios derived from the balanced chemical equation.

Gas Laws at STP

Standard Temperature and Pressure (STP) is a reference condition defined as 0°C (273.15 K) and 1.00 atm pressure, at which one mole of an ideal gas occupies a fixed volume (22.414 L). Familiarity with this concept allows for straightforward conversion between moles of gas and volume when reactions occur at STP.

Ideal Gas Law

The Ideal Gas Law, expressed as PV = nRT, relates the pressure, volume, temperature, and moles of a gas. This equation is essential when calculating the volume of a gas under conditions that differ from STP, as it provides a method to adjust for non-standard temperature and pressure.

Transcript

00:01 Okay, so in this question we have two parts, part a and b.

00:06 Let's start with part a first.

00:08 So let's write down the information that we're given in the question.

00:12 We're given a temperature of 0 degree celsius, which equals 273 kelvin.

00:17 We're given a pressure of one atmosphere, and we're given a mass of 8 .25 grams.

00:24 And this is the mass of magnesium metal.

00:27 We need to find the volume.

00:32 And so for this question, we can see that these variables are perfect for the ideal gas law.

00:42 So this is the ideal gas law in red.

00:45 We know the temperature.

00:46 We know the pressure.

00:48 We don't know the moles yet, but we can find the moles, which i will show you how to do in a little bit.

00:54 We always know the gas constant, and this question wants us to find the volume.

00:59 So there's two steps here.

01:01 First, we are going to find the moles of magnesium based on the mass we are given of magnesium.

01:18 Once we find the moles of magnesium, we can plug it in our ideal gas law to find the volume.

01:29 So let's go ahead and do step one first.

01:31 We're going to find the moles of magnesium.

01:33 So we're going to use our formula here in green.

01:36 N equals mass, moles, excuse me, equals mass over.

01:41 Molar mass.

01:43 Moles equals mass over molar mass.

01:45 So we're given the mass of 8 .25 grams.

01:49 We can find the molar mass by taking a look at our periodic table.

01:53 24 .31 grams per mole.

01:56 Molar mass of magnesium.

01:59 You should get about 0 .339 moles.

02:03 So this is our moles of magnesium.

02:10 Now because we know the moles, we have all the right.

02:15 Variables for the ideal gas law here in blue.

02:19 So we're just going to go ahead and plug in our numbers.

02:22 1 atm for your pressure, moles, we just found them.

02:27 0 .339 moles.

02:31 We know temperature 273 kelvin, and the gas constant is 0 .0821 liters times atmospheres over moles times k.

02:44 So this, if you rearrange and solve for volume, you should get about 7 .61 liters.

02:54 Oops, and this is your final answer for part a.

03:00 That is the volume.

03:02 So moving along to part b, we are given a few different variables here, and this question is a little bit different, but we'll walk you through it.

03:14 So in this question, we want to know the mass of magnesium this time.

03:19 Okay.

03:20 We are given the volume of hydrogen gas to be 5 liters.

03:26 So you have to keep track of what variable you're given for what part of the reaction, okay, for what compound.

03:34 So we're given a pressure of 735 millimeters of mercury, and we're going to go ahead and convert this into atmospheres.

03:43 Because you see that the gas constant has atmospheres in its unit...

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