Gas laws (Balancing equations)

Question

Calculate the volume of hydrogen measured at 91C and 80mmHg which is evolved when 12g og magnesium are treated with excess hydrochloric acid. Write a balanced equation

Answer 1

Seeing as how you are given Temperature, mass in grams, and pressure and you are asked to find volume,. You can use one law that relates all of these properties but still uses a gas constant. That is: PV=nRT.

You would first start off by writing a reaction of what this would look like, then entering the right amount of moles based on stoichiometric balance then in order to solve for the correct amount of volume for that moles.

You are told you have HCL in excess reacting with magnesium

Mg + HCl==> MgCl + H

This would have to be a replacement reaction. You would have to balance according to the charges:.

Mg has a +2 charge so

Mg+ HCl==> MgCl2. + H

Now balance everything and you have:

Mg+ 2HCl==> MgCl2 + H2 .

Now you have 1 Mg on each side

2H on each side

And 2 Cl on each side

Now based on this balanced equation you can use stoichiometry to know how many final moles you will have:

Since you are given this information, you are given 12g of magnesium, how many moles is this?

Mg atomic weight:. 24.305g/mol

.

12g * 1mol/24.305g. Grams cancel top to bottom and you have moles in the numerator.

0.4937moles of magnesium

Since magnesium is NOT in excess but instead the hydrochloric acid is:

1molMg==>1molH2

0.4937molMg==>0.4937molH2

X= 0.4937mol H2

Now use this information and plug it into the other equation

PV=nRT

1 atm = 760mmHg

.

80mmHg *'' 1atm/760mmHg, mmHg cancel top to bottom and you have atm in the numerator at:

0.10526atm

Temp 91C +273.15 = 364.15 K

Now you have all the right units, select the correct R constant reflecting these units:

0.0821 atm*L/mol*K

0.10526*V=0.4937(0.0821)364.15

Solve for V

V=(0.4937(0.0821)364.15)/(0.10526)

Units resolve and you have,

140.22 L

For 2 sig figs

140 L

Answer 2

First, write the correctly balanced equation for the reaction taking place:

Mg(s) + 2HCl(aq) ==> MgCl₂(aq) + H₂(g)

Next, find moles of Mg and then moles of H2 gas produced:

mols Mg = 12 g Mg x 1 mol Mg / 24 g = 0.5 mols Mg

mols H2 produced = 0.5 mols Mg x 1 mol H2 / mol Mg = 0.5 mols H2 produced

To find the volume of H2, use the ideal gas law, PV = nRT

P = pressure in atm = 80 mm Hg x 1 atm / 760 mm Hg = 0.105 atm

V = volume in L = ?

n = moles H2 = 0.5 mols

R = gas constant = 0.0821 Latm/Kmol

T = temperature in Kelvin = 91C + 273 = 364K

Solving for V, we have...

V = nRT/P = (0.5)(0.0821)(364) / 0.105

V = 142 L = 140 L (2 sig. figs.)