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#49  In 1897 the Swedish explorer Andree tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is given below.

Fe(s) + H2SO4(aq) -->FeSO4(aq) + H2(g).

The volume of the balloon was 4800 m3 and the loss of hydrogen gas during filling was estimated at 20.%. What mass of iron splints and 98% (by mass) H2SO4 were needed to ensure the complete filling of the balloon? Assume a temperature of 0C, a pressure of 1.0 atm during filling, and 100% yield.

 
Answer 

4800m3

=    X           X=6.0 x 103 m3 H2 produced

80%   

100%

Since it is at STP 22.4L = 1 mol of a gas

6.0 x 103 m3 H2

1,000 L  X 1 mol =2.7 x 105mol H2(g)

1 m3

22.4 L

Since the reactants are in a ratio with H2 of 1:1:1

moles Fe=moles H2SO4 =moles H2= 2.7 x 105mol

 Grams Fe=  2.7 x 105mol Fe x

55.85g Fe

=1.5 x 107g Fe 

1 mol Fe

Grams H2SO4=2.7 x 105mol H2SO4 x

98.08g H2SO4

=2.6 x 105g H2SO4

1 mol H2SO4

Since it is 98%

2.6 x 105g H2SO4

=  X g                   X=2.6 x 105g  98% H2SO4

98%              

100%

 

How many liters of hydrogen gas are collected over water at 26C and 725 mmHg when 0.73 g lithium reacts with water? Aqueous lithium hydroxide also forms.
 
Write the reaction                2Li + 2H20--> 2LiOH + H2(g)     so for every 2 moles of Lithium you yield 1 mole H2 gas
Let's Find the Moles of Lithium

0.73 g lithium

1 mole Li = 0.105mol Li
6.941g Li

Convert to moles H2        0.105mol Li x

1mole H2 =0.0.0525 mol H2
2 mole Li
Ideal Gas Law  PV= nRT

Convert 26C +273= 299K

 

Convert 725 mmHg X  

1 Atm = 0.954Atm
760mmHg
Let's Plug in the numbers
PV=nRT
(0.954Atm)(V) = (0.0.0525 mol H2)(0.0821L atm K-1 mol-1)(299K)

Solve for V =     (0.0525mol H2)(0.0821L atm K-1 mol-1)(299K)

= 1.35L

 (0.954Atm)                                  

Methanol, CH3OH, can be produced by the following reaction.
CO(g) + 2 H2(g)==> CH3OH(g)

Hydrogen at STP flows into a reactor at a rate of 15.0 L/min. Carbon monoxide at STP flows into the reactor at a rate of 25.0 L/min. If 5.75 g of methanol is produced per minute, what is the percent yield of the reaction?
Let's use the amount in 1 minute

Since it is at STP 22.4L =1 mole

15.0 L Hydrogen X 

1 mol H2

X

1 mol Methanol

X

32.0g Methanol =10.7 g Methanol/min
22.4 l H2

2 mole H2

1 mol Methanol

25.0 L Carbon monoxide X

1 mol CO

X

1 mol Methanol

X

32.0g Methanol = 35.7g Methanol/min
22.4 l CO

1 mol CO

1 mol Methanol
Since only 10.7g Methanol/min is the limiting yield (it is less)
% yield=   Measured x 100% = 5.75g.min x 100%=53.7%
Theoretical 10.7g.min

Urea (H2NCONH2) is used extensively as a nitrogen source in fertilizers. It is produced commercially from the reaction of ammonia and carbon dioxide.

2 NH3(g) + CO2(g) -->H2NCONH2(s) + H2O(g)

Ammonia gas at 223C and 90. atm flows into a reactor at a rate of 460. L/min. Carbon dioxide at 223C and 46 atm flows into the reactor at a rate of 600. L/min. What mass of urea is produced per minute by this reaction assuming 100% yield?
____________g/min

Answer

Let's take 1 minutes work of data

Ammonia

T=223C + 273=500K

P=90.atm

V=460.L

find moles of NH3                PV=nRT

(90.atm)(460. L)=n(0.0821L atm K-1 mol-1)(500K)

n=1.0 x 103 mols NH3

Carbon dioxide 

T=223C + 273=500K

P=46 atm

V=600. L

find moles of CO2                   PV=nRT

(46.atm)(600. L)=n(0.0821L atm K-1 mol-1)(500K)

n=6.7 x 102 mols CO2

1.0 x 103 mols NH3 X

1 mol Urea X 60 g Urea =3.0 x 103g Urea/min  Limiting (lesser amount)
2 mols NH 1 mol Urea

6.7 x 102 mols CO2 X

1 mol Urea X 60 g Urea =4.0 x 104g Urea/min
1 mols CO2 1 mol Urea

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