Synthesis gas stream consisting of 65.80% H2, 33.80% CO, and 0.400% CH4 (by mole) is fed into the process described below at a rate of 100 mol/hour. The effluent stream from the reactor is fed into a separator where the methanol is completely removed, and the unconverted reactants are recycled to the reactor. To avoid the buildup of methane, a portion of the recycled stream is purged. At present operating conditions, the CO conversion over the entire process is 92.00%, and the methane mole fraction at the reactor inlet (Stream 2) is 2.870%.

the answer would be b i believe

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