In an experiment, `6.67 g` of `AlCl_(3)` was produced and `0.54g` `Al` remained unreacted. How many g atoms of `Al` and `Cl_(2)` were taken originally?

Options

A. 0.070, 0.075
B. 0.07, 0.05
C. 0.02, 0.15
D. 0.02, 0.15

Answer

A

Explanation

`Al + (3)/(2)Cl_(2)rarr AlCl_(3)`
`{:("Initially",,),("Finally",(0.54)/(27) = 0.02 "moles",(6,67)/(133.5) "moles" = (1)/(20) "moles" = 0.05 "moles"):}`
`rArr` Initially there were 0.070 moles of Al and 0.075 moles of `Cl_(2)`