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Question: A mixture of hydrogen and oxygen 45ml is sparked to form liquid water. The component not in excess r...

A mixture of hydrogen and oxygen 45ml is sparked to form liquid water. The component not in excess reacts completely and 15ml15ml is left over. (All measurements are made at the same temperature and pressure). The composition by volume in the original mixture of H2:O2{H_2}:{O_2} is _____.
A. 4:54:5
B. 7:27:2
C. 2:12:1
D. either 4:54:5 or 7:27:2

Explanation

Solution

In the given question firstly we have to get both the conditions for the given statement where each of the gas would be left in each of the cases subsequently and individually. So we will get the concentration different and then get the two answers would be two for equation. Now we will get the two concentrations as H2  = 35 mL  {H_2}\; = {\text{ 35 }}mL\; ,   O2  = 10 mL\;{O_2}\; = {\text{ }}10{\text{ }}mL and H2  = 20 mL  {H_2}\; = {\text{ 20 }}mL\; ,   O2  = 25 mL\;{O_2}\; = {\text{ 25 }}mL .

Complete step by step answer:
Firstly we have to sort out details in the given problem, that are :
The given mixture of the gases is of the gases of : hydrogen and oxygen
The amount of the gases used would be : 45ml45ml
The amount of gases left over : 15ml15ml
Now firstly we have to give the reaction for the given process that is happening :
Reaction in which water is formed :
2H2+O22H2O2{H_2} + {O_2} \to 2{H_2}O
So we have to make the two cases where we would consider that each time the excess gas is left is different.
So for the cases:
Case 1: Here we take the excess gas left to be H2{H_2}
So the total amount of the gas used would be 45ml15ml=30ml45ml - 15ml = 30ml
H2  = 20 mL  {H_2}\; = {\text{ }}20{\text{ }}mL\;
  O2  = 10 mL\;{O_2}\; = {\text{ }}10{\text{ }}mL
So that will give us the original concentration as
H2  = 35 mL  {H_2}\; = {\text{ 35 }}mL\;
  O2  = 10 mL\;{O_2}\; = {\text{ }}10{\text{ }}mL
So the ratio would be :
=35:10 =7:2  = 35:10 \\\ = 7:2 \\\
​Case 2: Here we take the excess gas left to be O2{O_2}
So the total amount of the gas used would be 45ml15ml=30ml45ml - 15ml = 30ml
H2  = 20 mL  {H_2}\; = {\text{ }}20{\text{ }}mL\;
  O2  = 10 mL\;{O_2}\; = {\text{ }}10{\text{ }}mL
So that will give us the original concentration as
H2  = 20 mL  {H_2}\; = {\text{ 20 }}mL\;
  O2  = 25 mL\;{O_2}\; = {\text{ 25 }}mL
So the ratio would be :
=20:25 =4:5  = 20:25 \\\ = 4:5 \\\
Therefore the most correct option would be option D, either 4:54:5 or 7:27:2.

Note:
Water is an inorganic, transparent, tasteless, odorless, and nearly colorless chemical substance, which is the main constituent of Earth's hydrosphere and the fluids of all known living organisms. Formation waters are mixtures of original trapped fluids such as seawater and fluids modified by rock–water interaction (hydration or dehydration reactions or crustal transport).