Question

A mixture of 50.0 mL of ammonia gas and 60.0 mL of oxygen gas react. If all the gases are at the same temperature and pressure, and the reaction continues till one of the gases is completely consumed, what volume of steam is produced if the other product is nitric oxide gas (NO).

Answer 1

The volume of steam produced: 72 ml

Further explanation

Some of the laws regarding gas can apply to ideal gas (volume expansion does not occur when the gas is heated),:

• Boyle's law at constant T,

$\displaystyle P = \dfrac {1} {V}$

• Charles's law, at constant P,

$\displaystyle V = T$

• Avogadro's law, at constant P and T,

$\displaystyle V = n$

So that the three laws can be combined into a single gas equation, the ideal gas equation

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

A mixture of 50.0 mL of ammonia gas and 60.0 mL of oxygen gas reacts at the same temperature and pressure, then we use Avogadro's law:

V ≅ n

So the reaction coefficient (showing the mole ratio) that occurs is proportional to the amount of gas volume reacting

Reaction

 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

This reaction is one of the processes of making nitric acid through the Oswald process

A method that can be used to find limiting reactants is to divide the number of volumes of known reactants by their respective coefficients, and small or exhausted reactants become a limiting reactants

NH₃: volume: reaction coefficient = 50: 4 = 12.5

O₂: volume: reaction coefficient = 60: 5 = 12

So from this comparison, O₂ has the smallest ratio so that it becomes a gas whose volume has completely reacted  (become a limiting reactants )

Then the determination of the volume of steam (H₂O) is based on the volume of O₂

Reaction coefficient H₂O: O₂ = 6: 5

then the volume of steam (H2O):

$\rm \dfrac{6}{5}\times 60=\boxed{\bold{72\:ml}}$

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Answer 2

Answer:

72.0 mL of steam is formed.

Explanation:

The reaction is :

$4 NH_{3} + 5O_{2} \rightarrow 4NO+6H_{2} O$

You can treat coefficient of compounds as amount of volume used.

Therefore for 4 mL of ammonia 5 mL of oxygen is used to form 4 mL of nitric oxide gas and 6 mL of steam.

For 1 mL of ammonia $\frac{5}{4}$ (=1.25) mL of oxygen is used to form $\frac{4}{4}$ (=1) mL of nitric oxide gas and $\frac{6}{4}$ (=1.5) mL of steam.

OR

Just transform the chemical equation by dividing the whole equation by 4 so that the coefficient of $NH_{3}$ become one like this

$NH_{3} + \frac{5}{4} O_{2} \rightarrow \frac{4}{4}NO+\frac{6}{4}H_{2} O$

We don't know which one will be completely exhausted and which one will be left so we have to consider two cases :

1.  Assume ammonia to be completely exhausted

For 50 mL of ammonia $50 \times \frac{5}{4}$ (= 62.5) mL of oxygen is needed. But we have just 60 mL of oxygen so this assumption is false.

2.  Assume oxygen to be completely exhausted

For 60 mL of oxygen only $60 \times \frac{4}{5}$ (=48) mL of ammonia is needed. In this case we have sufficient amount of ammonia. So this case is true.

$60\times\frac{4}{5}NH_{3} + 60\ O_{2} \rightarrow 60\times\frac{4}{5}NO+60\times\frac{6}{5}H_{2} O\\\\48NH_{3} + 60\ O_{2} \rightarrow 48NO+72H_{2} O$

Now we know that during complete reaction 48 mL of ammonia and 60 mL of oxygen is used which will form $60 \times \frac{4}{5}$ (= 48) mL of nitic oxide gas and $60 \times \frac{6}{5}$ (= 72) mL of steam.

Therefore 72 mL of steam is formed.