Compound x, c14h12br2, is optically active. on treatment with strong base, compounds k and l (each c16h14) are formed; k and l each absorb only 2 equivalents of hydrogen when reduced over a pd/c catalyst. compound k reacts with ozone to give phenylacetic acid (c6h5ch2cooh), while similar treatment of l gives 2 products. one product, m, is an aldehyde with formula c7h6o; the other product is glyoxal (cho)2. draw the structure of compound m.

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