The specific rotation of (S)-2- butanol is + 13.5. If 1.00 g is dissolved in 10.0 mL of ethanol and placed in a sample cell with a length of 1.00 dm, what observed rotation do you expect?

-1.35 degree

Explanation:

Given the specific rotation of (S)-2-butanol = +13.5 degree

So, the specific rotation of its enantiomer (R)-2-butanol = -13.5 degree

Now the optical rotation of a pure liquid is given by

Here,  = the optical rotation

          = specific rotation at temperature 't' and wave length 'λ'

                = -13.5 degree

         l = optical path length

           = 1 dm

          d  = density

             = 1/10

Therefore,

    = -1.35 degree

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