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Question: When \( 10ml \) of \( {H_2} \) and \( 12.5ml \) of \( C{l_2} \) are allowed to react, the mixture co...

When 10ml10ml of H2{H_2} and 12.5ml12.5ml of Cl2C{l_2} are allowed to react, the mixture contains under the same conditions.

Explanation

Solution

Hint : in this particular reaction we have to assume that the reaction is taking place under standard conditions i.e. at standard temperature and pressure. If not, this question has two ways to be solved: first one is considering STP and the second method is just by looking at the values given in mlml .

Complete Step By Step Answer:
At STP conditions, 10ml10ml of hydrogen will correspond to 0.450.45 millimoles and 12.5ml12.5ml of chlorine will correspond to 0.550.55 millimoles respectively.
The reaction that will take place here is: H2+Cl2→2HCl{H_2} + C{l_2} \to 2HCl
As we can see in the above reaction that 1 mole of hydrogen and 1 mole of Chlorine react with each other to give 2 moles of HCl.
And so 0.450.45 millimoles of hydrogen will react with 0.450.45 millimoles of Chlorine and will give 0.90.9 millimoles of HClHCl .
So the amount of Chlorine that remains unreacted in the reaction is 0.10.1 millimoles.
Therefore, the reaction mixture will contain 0.100.10 millimoles of chlorine and 0.90.9 millimoles of HCl.
And hence at STP this will correspond to 20.16 ml20.16{\text{ }}ml of HCl and 2.24 ml2.24{\text{ }}ml of Chlorine.

Note :
This problem can be solved alternately also. For solving it in a different way we do not consider any STP, by using the mlml of reactant given we solve this problem.
The reaction that will take place here is: H2+Cl2→2HCl{H_2} + C{l_2} \to 2HCl
And we are given with 10ml10ml of hydrogen and 12.5ml12.5ml of chlorine
1ml1ml Of H2{H_2} will react with 1ml1ml of Cl2C{l_2} to give 2ml2ml of HClHCl .
And so 10ml10ml of H2{H_2} will react with 10ml10ml of Cl2C{l_2} to give 20ml20ml of HClHCl .
Therefore the amount of chlorine that is left in the mixture:
= 12.5ml - 10ml \\\ = 2.5ml \\\
And the total amount of HClHCl formed is 20ml20ml .
As you can see , solving with this method also gets almost the same values as that of the first method. There is not much difference so this is also acceptable. To if one needs to find it in a simple way this method can be taken into consideration.