In an experiment, `6.67g" of "AlCl_(3)` was produced and `0.54 g Al` remained unreacted. How many g atoms of Al and `Cl_(2)` were taken orginally (Al= 27, Cl = 35.5)?

Options

A. `0.07, 0.15`
B. `0.07,0.05`
C. `0.02, 0.05`
D. `0.02, 0.15 `

Answer

A

Explanation

A moles of `Al Cl _(3)` produced `= ( 6.67)/(133.5) = 0.05 ` mol
Number of moles of axcess of `Al = ( 0.54)/(27) = 0.02 mol`
g atom or moles of Al taken `= 0. 05 + 0.02 = 0.07`
g atom of moles of `Cl _(2)` taken `= 3 xx 0.05 = 0.15`