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?

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) ?

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) ?

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)?

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

In Al_2Cl_6 each Al atoms is linked to how many Cl atoms ?

A mixture of 1.0 mole of Al and 3.0 mole of Cl_(2) are allowed to react as: 2Al_((s))+3Cl_(2) rarr 2AlCl_(3(S)) (a) Which is limiting reagent? (b) How many moles of AlCl_(2) are formed? (c ) Moles of excess reagent left unreacted.