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

Question

In an experiment, 6.67 g of AlCl 3 was produced and 0.54 g Al remained 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

Tardigrade Admin · Accepted Answer

Moles of AlCl 3 produced =(6.67/133.5)=0.05 mol Excess of Al =(0.54/27)=0.02 mole g atom or moles of Al taken =0.05+0.02=0.07 g atom or moles of Cl 2 taken =3 × 0.05=0.15

Q. In an experiment, $6.67 g$ of $A lC l_{3}$ was produced and $0.54 g A l$ remained unreacted. How many $g$ atoms of $A l$ and $C l_{2}$ were taken originally $(A l = 27, Cl = 35.5)$ ?

$0.07, 0.15$

$0.07, 0.05$

$0.02, 0.05$

$0.02, 0.15$

Moles of $A lC l_{3}$ produced $= \frac{6.67}{133.5} = 0.05 m o l$
Excess of $A l = \frac{0.54}{27} = 0.02$ mole
$g$ atom or moles of $A l$ taken $= 0.05 + 0.02 = 0.07$
$g$ atom or moles of $C l_{2}$ taken $= 3 \times 0.05 = 0.15$

Q. In an experiment, 6.67g of AlCl3​ was produced and 0.54gAl remained unreacted. How many g atoms of Al and Cl2​ were taken originally (Al=27,Cl=35.5) ?

0.07,0.15

0.07,0.05

0.02,0.05

0.02,0.15