If 1.06mmole of 1-pentanol and 1.53mmole of 1-Hexanol were separated by gas chromotography, they gave relative peak areas of 922 and 1570 units, respectively. When an unknown mixture of these two alcohols was likewise separated, the relative peak areas were 843:816 (pentanol:hexanol). What was the relative concentration of these species in the mixture?

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the answer is true. hope it .

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