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In an experiment, 6.67 g of AlCl(3) was ...

In an experiment, `6.67 g` of `AlCl_(3)` was produced and `0.654 g` Al remainded unreacted. How many `g` atoms of `Al` and `Cl_(2)` were taken originally `(Al = 27, Cl = 35.5)`?

A

0.07,0.15

B

0.07,0.05

C

0.02,0.05

D

0.02,0.15

Text Solution

Verified by Experts

The correct Answer is:
A
Moles of `AlCl_(3)` porduced `= (6.67)/(133.5) = 0.05` mol
Excess of `Al = (0.54)/(27) = 0.02` mole
`g` atom of moles of `Al` taken `= 0.05 + 0.02 = 0.07`
`g` atom of moles of `Cl_(2)` taken `= 3 xx 0.05 = 0.15`
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