How many moles of SbCl3 are formed when 4.00 mol Sb reacts 4.43 mol of Cl2 according to the following unbalanced equation Sb plus Cl2 --- SbCl3?

Answer 1

First, balance the equation:2Sb + 3Cl2 ==> 2SbCl3

Next, figure out which reactant is limiting:

4 mol Sb/2 = 2

4.43 mol Cl2/3 = 1.48 => Cl2 is LIMITING

Finally, calculate moles of SbCl3 produced from 1.48 moles Cl2:

1.48 moles Cl2 x 2 moles SbCl3/3 moles Cl2 = 0.987 moles SbCl3 formed

Answer 2

2.95 mol SbCl3 can be produced in this reaction.

This is a limiting reagent (reactant) problem. The maximum amount of antimony trichloride (SbCl3) that can be produced is determined by the limiting reagent.

First balance the equation:

2 Sb + 3 Cl2 --> 2 SnCl3

Multiply the given mol of each reactant by its mol ratio with SbCl3 in the balanced equation, so that SbCl3 is in the numerator.

4.00 mol Sb x (2 mol SbCl3)/(2 mol Sb) = 4.00 mol SbCl3

4.43 mol Cl2 x (2 mol SbCl3)/(3 mol Cl2) = 2.95 mol SbCl3

The limiting reagent is chlorine gas (Cl2). The maximum amount of SbCl3 that can be produced is 2.95 mol SbCl3.