Q1)Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)2A

Question

Q1)Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)2AlCl3(s).What minimum volume of chlorine gas (at 298 K and 229 mmHg ) is required to completely react with 8.60 g of aluminum? please explain the step, how you got to the answer for this problem.

A)38.8 mL

B)38.8 L

C)388 L

D)0.388 L

Q2)Calculate the density of oxygen, O2 under each of the following conditions

STP and 1.00atm and 35.0 Celcius

(Express answer numerically in grams per liter. Enter density at STP first & seperate your answer by a comma)

Q3) To identify a diatomic gas(

X2), a researcher carried out the following experiment: She weighed an empty 4.3-Lbulb, then filled it with the gas at 1.90 atm and 23.0 Cand weighed it again. The difference in mass was 9.5 g . Identify the gas.(Express answer as chemical formula)

Explanation / Answer

Solution : -

Q1)Aluminum reacts with chlorine gas to form aluminum chloride. 2Al(s)+3Cl2(g)2AlCl3(s).What minimum volume of chlorine gas (at 298 K and 229 mmHg ) is required to completely react with 8.60 g of aluminum?

Balanced reaction equation

2Al + 3Cl2 ----- > 2AlCl3

Now lets calculate the moles of 8.62 g Al

8.60 g Al * 1 mol /26.982 g per mol = 0.31873 mol Al

Now lets calculate moles of Cl2 using the moles of Al

0.31873 mol Al * 3 mol Cl2 / 2 mol Al = 0.4781 mol Cl2

Now lets calculate the volume of the Cl2 gas needed at the given conditions

PV= nRT

V= nRT/P

= 0.4781 mol * 0.08206 L atm per mol K * 298 K / (229 mmHg * 1 atm / 760 mmHg)

= 38.8 L

So the volume of Cl2 needed is 38.8 L

Q2)Calculate the density of oxygen, O2 under each of the following conditions

STP and 1.00atm and 35.0 Celcius

Formula to calculate the density is

PM= dRT

d=PM/RT

= 1 atm * 32.0 g per mol / 0.08206 L atm per mol K * (35.0 C +273 )

= 1.226 g/ L

So the density of the oxygen gas is 1.226 g / L

Solution :-

Lets first calculate the moles of the gas

P = 1.90 atm

T= 23.0 C +273 = 296 K

V = 4.3 L

PV= nRT

PV/RT = n

1.90 atm * 4.3 L / 0.08206 L atm per mol K * 296 K = n

0.33636 mol = n

Now using the moles and mass of the gas lets calculate its molar mass

Molar mass = mass / moles

                     = 9.5 g / 0.33636 mol

                     = 28.2 g per mol

So the diatomic gas is N2 because molar mass of N2 = 28.01 g per mol

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