Question

with
11.2 litres
11
When 22.4 litres of H2(g) is mixed with
C each at STP, the moles of HCl(a) f
(g) formed is equal to
ICBSF Pm​

Answer 1

Answer:

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