Question

$3d{{m}^{3}}$ of ${{H}_{2}}$ is mixed with \[4d{{m}^{3}}\] of \[C{{l}_{2}}\] at NTP. Calculate volume of HCl obtained and volume of \[C{{l}_{2}}\] unreacted.

Answer 1

Hint: NTP stands for Normal Temperature and Pressure and the term is defined by the National Institute of Standards and Technology abbreviated as NIST. NTP conditions are more realistic and convenient conditions so it is widely used as a practical standard by Industries throughout the world.

Complete step by step answer:
- In NTP the normal temperature is 20-degree Celsius, 293.15 K or 68-degree F. This temperature is convenient to use as compared to 0-degree Celsius , normal pressure is 1 atm,101.325 kpa, 760 mmHg (or torr) or 14.6959 psi and one mole of an ideal gas occupies 24.0 litres ($d{{m}^{3}}$) of volume.
According to the question:
$3 d{{m}^{3}}{{H}_{2}}+4 d{{m}^{3}}C{{l}_{2}}\to xHCl$
Suppose it gives x $d{{m}^{3}}$ of HCl.
Now we know that normally 1 $d{{m}^{3}}$ of ${{H}_{2}}$ reacts with 1 $d{{m}^{3}}$ of $C{{l}_{2}}$ and make 2 $d{{m}^{3}}$ of HCl. Reaction can be shown as follows:
${{H}_{2}}+C{{l}_{2}}\to 2HCl$
So from here we can conclude that ${{H}_{2}}$ acts as limiting reagent.
So we can consider that 3 $d{{m}^{3}}$ of ${{H}_{2}}$ reacts and make 6 $d{{m}^{3}}$ of HCl and in this case 1 $d{{m}^{3}}$ of $C{{l}_{2}}$ remain unaltered.
Hence volume of HCl obtained is 6 $ d{{m}^{3}}$ and volume of \[C{{l}_{2}}\] unreacted is 1 $d{{m}^{3}}$.

Note: Limiting reagents are defined as those substances which are completely consumed in the completion of a chemical reaction and can also be known by the name limiting agents or limiting reactants. According to the stoichiometry of chemical reactions, a fixed amount of reactants is required for the completion of the reaction.