Question

please someone answer it in detail

Answer 1

C. 108.0 g

Since 22400 mL volume is occupied by 1 mole of O2 STP

Thus 5600 mL of O2 means =

(5600 ÷ 22400) mol O2

= 1/4 mol O2

Therefore, weight of O2 = (1/4) × 32 = 8 g

According to problem,

Equivalent of Ag = Equivalent of O2

= (weight of Ag ÷ equivalent weight of Ag)

= ( weight of O2 ÷ equivalent weight of O2)

W Ag/ M Ag = W O2/ M O2

Therefore, (W Ag/108) × 1 = (8/32) × 4

[2H2O --> O2 + 4H+ + 4e- ]

Therefore, W Ag = 108 g