If a sample contains 84.0 % of the R enantiomer and 16.0 % of the S enantiomer, what is the enantiomeric excess of the mixture

enantiomeric excess = 68%

Explanation:

Enantiomeric excess is a value used to determine the purity of chiral molecules. It is possible to determine enantiomeric excess (ee) using:

ee = R - S / R + S * 100

Where R is the mass (In this case percentage) of the R enantiomer and S of the S enantiomer.

Replacing with values of the problem:

ee = 84% - 16% / 84% + 16% * 100

what is the question here. i think you messed up.

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