Question 3.82: Count the number of each type of molecule in the reactant bo......
Count the number of each type of molecule in the reactant box and in the product box. Subtract any molecules of excess reagent (molecules appearing in both boxes). The remaining material is the overall equation.
This will need to be simplified if there is a common factor among the substances in the equation. The balanced chemical equation is necessary for the remainder of the problem.
a) The contents of the boxes give:
6\text{ AB}_2 + 5\text{ B}_2 → 6\text{ AB}_3 + 2\text{ B}_2
\text{B}_2 is in excess, so two molecules need to be removed from each side. This gives:
6\text{ AB}_2 + 3\text{ B}_2 → 6\text{ AB}_3
Three is a common factor among the coeffic ients, and all coefficients need to be divided by this value to give the final balanced equation :
2\text{ AB}_2 + \text{B}_2 → 2\text{ AB}_3
b) \text{B}_2 was in excess, thus \text{AB}_2 is the limiting reactant.
\text{AB}_2 is the limiting reagent and maximum is 5.0 mol \text{AB}_3.
\text{d) Moles of B}_2\text{ that reacts with 5.0 mol AB}_2=(5.0\text{ mol AB}_2)\left\lgroup\frac{1\text{ mol B}_2}{2\text{ mol AB}_2} \right\rgroup =2.5\text{ mol B}_2The un-reacted \text{B}_2 is 3.0 mol – 2.5 mol = 0.5 mol \text{B}_2.