When 22.4 L of H2 (g) is mixed with 11.2 L of Cl2 (g), each at STP, the moles of HCl(g) formed is equal to
The given problem is related to the concept of stiochiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then, identify the limiting reagent [possessing minimum number of moles and gets completely used up in the reaction].
The limiting reagent gives the moles of product formed in the reaction.
H2(g) + Cl2(g) 2HCl (g)
Initial vol 22.4 L 11.2 L 2 mol
... 22.4 L volume at STP is occupied by,
Cl2 = 1 mole,
... 11.2 L volume will be occupied by,
Cl2 = 1x 11.2/22.4 mole = 0.5 mol
Thus, H2(g) + Cl2(g) 2HCl (g)