Gas Stoichiometry

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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 0°C, 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 X	1,000 L	X	1 mol	=2.7 x 10⁵mol H₂(g)
	1 m³	X	22.4 L	=2.7 x 10⁵mol H₂(g)

Since the reactants are in a ratio with H₂of 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
Grams Fe= 2.7 x 10⁵mol Fe x	1 mol Fe	=1.5 x 10⁷g 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 26°C 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 x	1 mole Li	= 0.105mol Li
	6.941g Li
Convert to moles H₂ 0.105mol Li x	1mole H₂	=0.210 mol H₂
	2 mole Li

Ideal Gas Law PV= nRT
Convert 26°C +273= 299K
Convert 725 mmHg X	1 Atm	= 0.954Atm
	760mmHg

Let's Plug in the numbers
PV=nRT
(0.954Atm)(V) = (0.210 mol H₂)(0.0821L · atm · K^-1 · mol^-1)(299K)
Solve for V = (0.210 mol H₂)(0.0821L · atm · K^-1 · mol^-1)(299K)	= 5.40L
(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

1 mol Methanol

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

1 mol Methanol

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 223°C and 90. atm flows into a reactor at a rate of 460. L/min. Carbon dioxide at 223°C 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=223°C + 273=500K P=90.atm V=460.L find moles of NH₃PV=nRT (90.atm)(460. L)=n(0.0821L · atm · K^-1 · mol^-1)(500K) n=1.0 x 10³mols NH₃		Carbon dioxide T=223°C + 273=500K P=46 atm V=600. L find moles of CO₂PV=nRT (46.atm)(600. L)=n(0.0821L · atm · K^-1 · mol^-1)(500K) n=6.7 x 10²mols CO₂
1.0 x 10³mols NH₃X	1 mol Urea	X	60 g Urea	=3.0 x 10³g Urea/min Limiting (lesser amount)
	2 mols NH₃		1 mol Urea
6.7 x 10²mols CO₂ X	1 mol Urea	X	60 g Urea	=4.0 x 10⁴g Urea/min
	1 mols CO₂		1 mol Urea