Question

When 22.4 litres of H2 (g) is mixed with 11.2 liters of Cl2 (g), each at STP, the moles of HCl(g) formed is equal to
(a) 1 mol of HCl (g)
(b) 2 mol of HCl (g)
(c).0.5 mol of HCl(g)
(d) 1.5 mol of HCl (g)

Answer 1

hi friends

H2 + CL2-> 2HCL
22.4 L 22.4 L 2×22.4

22.4 Litres hydrogen reacts with 22.4 litre of chlorine to give 44.8 L of HCL
the given volume of chlorine is 11.2 L 
here chlorine is the liming reagents as its . first consumed 
22.4 Litre of chlorine gives 73 litre of HCL 
so , 11.2 l gives 44.8/22.4 × 11.2 = 22.4 of HCL is formed 
that means no. of mole = 1 
because 1 mole of any gases occupies 22.4 L at STP

the answer is A . 1 mole of HCL (g)

thankyou

Answer 2

Answer: ANS IS 1 MOL OF HCL

Explanation:

GIVEN IN ATTACHMENT

Here the limiting reagent is chlorine as its fully consumed