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0.1 Analyzing chemical equations  (Page 2/3)

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Reaction involving gas

In analyzing chemical reaction involving gas, we make the assumption that gases are ideal gases. In general, there are three different situations in which we may use gas volume in chemical analysis :

  • Gas volumes are given at STP.
  • Gas volumes are given at other temperature and pressure condition.
  • Only gas volumes are involved in calculation

In certain situation, reaction involving gas enables us to use gas volumes itself (not the moles) to analyze the reaction. We can extend the concept of molar proportions to volume proportions directly. Such is the case, when only gas volumes are involved in the calculation.

Gas volumes are given at stp

For analyzing gas volumes at STP, we make use of Avogadro’s hypothesis. A volume of 22.4 litres of ideal gas contains 1 mole of gas. The number of moles present in a volume “V” at STP is :

n = V 22.4

Problem : Determine the amount of concentrated H 2 S O 4 required to neutralize 20 litres of ammonia gas at STP.

Solution : The chemical reaction involved is :

2 N H 3 + H 2 S O 4 → N H 4 2 S O 4

Applying mole concept :

2 moles of N H 3 ≡ 1 mole of H 2 S O 4

2 X 22.4 litres of N H 3 ≡ 98 gm of H 2 S O 4

Amount of H 2 S O 4 for 20 litres of N H 3 = 98 44.8 X 20 = 43.8 g m

Gas volumes are given at other temperature and pressure condition

Under this condition, we first need to convert gas volumes to volumes at STP. We make use of ideal gas law,

P 1 V 1 T 1 = P 2 V 2 T 2

We may specify one of the suffix like “1” to represent the given condition and suffix “2” to represent the STP condition.

Problem : Determine the amount of magnesium required to liberate 900 cc of hydrogen from a solution of HCl at 27°C and 740 mm of Hg. A M g = 24 .

Solution : The chemical reaction involved is :

M g + 2 H C l → M g C l 2 + H 2

Applying mole concept :

1 mole of Mg ≡ 1 mole of H 2

24 gm of Mg ≡ 22400 cc of H 2 at STP

Using ideal gas law,

⇒ V 2 = P 1 V 1 T 2 P 2 T 1 = 740 X 900 X 273 760 X 300 = 797 c c

Using this data to mole relation,

Amount of Mg = 24 X 797 22400 = 0.85 g m

Only gas volumes are involved in calculation

In this case, we are required to consider volumes of gases only. What it means that there is no solid or liquid elements involved in calculation. The reaction may involve solid or liquid components but they are not involved in calculation. In such situation, we can connect volumes of gas components of the reaction directly. Consider the reaction :

H 2 + C l 2 → 2 H C l

x litres of H 2 ≡ x litres of C l 2 ≡ 2x litres of H C l

This direct relation is deducted from mole concept under two assumptions (i) all gases are ideal and (ii) temperature and pressure conditions are same for measuring all gases. Under these assumptions, equal numbers of moles of ideal gases occupy same volume. Hence, we can correlate volumes directly as above.

Problem : A 60 cc mixture of N 2 O and NO is mixed with excess H 2 and the resulting mixture of gas is exploded. The resulting mixture contains 38 cc of N 2 . Determine the volume composition of the original mixture.

Solution : The chemical reactions involved are :

N 2 O + H 2 → N 2 + H 2 O

2 N O + 2 H 2 → N 2 + 2 H 2 O

Let volume of N 2 O be x in the original mixture. Hence, volume of NO in the sample is 60 – x. Now, the corresponding mole relations are :

x cc of N 2 O ≡ x cc of N 2

(60−x) cc of N 2 O ≡ 60 − x 2 cc of N 2

According to question,

x + 60 − x 2 = 38

x + 60 = 76

⇒ Volume of N 2 O = x = 16 c c

⇒ Volume of NO = 60 − x = 60 − 16 = 44 c c

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Source:  OpenStax, Stoichiometry. OpenStax CNX. Jul 05, 2008 Download for free at http://cnx.org/content/col10540/1.7
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