Styrene (1.53 g, 0.011 mol) in methanol (30 mL) was added to a mixture of Hg(OAc)2 (5.30 g, 0.016 mol) in methanol (100 mL) at room temp. and stirred for 24 h. Sodium hydroxide (3.0 M, 16 mL) was added, followed by NaBH4 (0.32 g, 0.008 mol) in NaOH (3.0 M, 16 mL) at 0 °C. The precipitated Hg was removed by filtration. The product was isolated by diethyl ether extraction. After drying over Na2SO4, solvent was removed and distillation gave the product. (Adapted from: Senda, Y.; Kanto, H.; Itoh, H. J. Chem. Soc., Perkin Trans. 2 1997, 1143-1146.)

i think its 5049.

i hope this !

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