A 7.30-L container holds a mixture of two gases at 47 °C. The partial pressures

Question

A 7.30-L container holds a mixture of two gases at 47 °C. The partial pressures of gas A and gas B, respectively, are 0.298 atm and 0.568 atm. If 0.170 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

A sample of an ideal gas has a volume of 3.35 L at 12.20 °C and 1.60 atm. What is the volume of the gas at 18.80 °C and 0.988 atm?

An ideal gas in a sealed container has an initial volume of 2.65 L. At constant pressure, it is cooled to 18.00 °C where its final volume is 1.75 L. What was the initial temperature?

If the initial temperature of an ideal gas at 2.250 atm is 62.00 °C, what final temperature would cause the pressure to be reduced to 1.650 atm?

A flexible container at an initial volume of 8.15 L contains 7.51 mol of gas. More gas is then added to the container until it reaches a final volume of 14.3 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.

Container A holds 742 mL of ideal gas at 2.30 atm. Container B holds 154 mL of ideal gas at 4.10 atm. If the gases are allowed to mix together, what is the resulting pressure?

How many grams of chlorine gas are needed to make 6.60 × 106 g of a solution that is 1.60 ppm chlorine by mass?

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation ,How much MnO2(s) should be added to excess HCl(aq) to obtain 315 mL of Cl2(g) at 25 °C and 775 Torr?

Explanation / Answer

a)

Moles of two gases A and B before addition of the third gas can be calculated from ideal gas law equation PV= nRT , n= PV/RT, P= total pressure = sum of partial pressures = 0.298+0.568=0.896 atm, V= 7.3 L, T= 47 deg.c = 47+273.15 K=320.15K, R= 0.0821 L.atm/mole.K

Number of moles = PV/RT=0.896*7.3/(0.0821*320.15)= 0.25 moles

When 0.170 moles are added, the total number of moles becomes =0.25+0.17=0.42

From PV= nRT, P= nRT/V= 0.42*0.0821*320.15/7.3=1.512 atm

b)

From P1V1/T1= P2V2/T2

P1= 1.6 atm V1= 3.35L, T1= 12.2 deg.c =12.2+273.15=285.35 K, P2= 0.988 atm and T2= 18.8deg.c =18.8+273.15=291.95 K

V2= P1V1*T2/P2T1= 1.6*3.35*291.95/ (0.988*285.35)=5.55 L

3.

Since the pressure remains constant, the ideal gas equation becomes

V1/T1= V2/T2 , where V1= 2.65 L, V2= 1.75L, T2= 18deg.c =18+273.15= 291.15K

T1= V1T2/V2= 2.65*291.15/ 1.75 =440.88K=167.73 deg.c

4.

T1= 62 deg.c = 62+273.15= 335.15 K, P1=2.25 atm P2= 1.65 atm

From P1V1/T1= P2V2/T2 . Assuming V1=V2

T2= P2T1/P1= 1.65*335.15/2.25 =245.7 K =-27.45 deg.c

5.

from PV= nRT since P and T remain constant

V1/n1= V2/n2

8.15/7.51= 14.3/n2

And hence n2=14.3*7.51/8.15=13.17 moles

Moles present initially= 7.51, moles additionally added = 13.17- 7.51=5.66 moles

6.

Since the gases are mixed , total volume = 742+154=896ml = 0.896L. Since the gas occupies all the available volume

P1V1= P2V2

2.3*0.742L= P2*0.896, P2=2.3*0.742/0.896=1.90 atm

The second gas at 4.1 atm and 154 ml

P2= 4.1*0.154/0.896=0.704 atm

7.

1.6 ppm =1.6 gm of chlorine/106 gm of solutiom

106 gm of solution contains 1.6 gm of chlorine

6.6*106 gm of solution contains 6.6*106*1.6/106 =10.56 gm of chlorine

8.

Moles of chlorine can be calculated from n= PV/RT

Where P= 775 Torr= 775/760 atm=1.019 atm, V= 315ml =0.315L, R= 0.0821 l.atm/mole.K T= 25 deg.c= 25+273.15= 298.15K

Number of moles, n= 1.019*0.315/(0.0821*298.15)=0.013 moles

Mass of chlorine =0.013*36.5=0.6045 gms

The reaction of MnO2 with HCl can be written as

MnO2 +4HCl --------à MnCl2 +Cl2 +2H2O

Moles of MnO2 required( since HCl is excess) = moles of Cl2 formed = 0.013 moles

Molecular weight of MnO2= 87, mass of MnCl2= 0.013*87=1.131 gms

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