. If sulphur (IV) oxide and methane are
released simultaneously at the opposite
ends of a narrow tube, the rates of
diffusion Rso, and Rch, will be in the
ratio (s = 32, O = 16, C = 12, H = 1]
(a) 4:1
(b) 2:1
(c) 1:2
(d) 1:4
hon

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2