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## Q. 3.10

Consider the reaction

2Sb(s) + 3$I_{2}$(s) → 2Sb$I_{3}$(s)

Determine the limiting reactant and the theoretical yield when

ⓐ 1.20 mol of Sb and 2.40 mol of $I_{2}$ are mixed.

ⓑ 1.20 g of Sb and 2.40 g of $I_{2}$ are mixed. What mass of excess reactant is left when the reaction is complete?

 ANALAYSIS moles of each reactant: Sb (1.20), $I_{2}$ (2.40) balanced equation: [2Sb(s) + 3$I_{2}$(s) → 2Sb$I_{3}$(s)] Information given: limiting reactant theoretical yield Asked for:

STRATEGY

1.  Find mol Sb$I_{3}$ produced by first assuming Sb is limiting, and then assuming $I_{2}$ is limiting.
mol Sb → mol Sb$I_{3}$; mol $I_{2}$ → mol Sb$I_{3}$
2. The reactant that gives the smaller amount of Sb$I_{3}$ is limiting, and the smaller amount of Sb$I_{3}$ is the theoretical yield.

 ANALAYSIS mass of each reactant: Sb (1.20 g), $I_{2}$ (2.40 g) balanced equation: [2Sb(s) + 3$I_{2}$(s) → 2Sb$I_{3}$(s)] Information given: molar masses (MM) of Sb$I_{3}$ and $I_{2}$ Information implied: limiting reactant theoretical yield mass of excess reactant not used up Asked for:

STRATEGY

1.  Follow the plan outlined in Figure 3.8 and convert mass of Sb and mass of $I_{2}$ to mol Sb$I_{3}$.
2. The smaller number of moles Sb$I_{3}$ obtained is the theoretical yield. The reactant that yields the smaller amount is the limiting reactant.
3. Convert moles limiting reactant to mass of excess reactant. That is the mass of excess reactant consumed in the reaction.
4. Mass of excess reactant not used up = mass of excess reactant initally – mass excess reactant consumed

## Verified Solution

 1.20 mol Sb × $\frac{2 mol SbI_{3}}{2 mol Sb}$ = 1.20 mol          2.40 mol $I_{2}$ × $\frac{2 mol SbI_{3}}{2 mol SbI_{3}}$ = 1.60 mol mol Sb$I_{3}$ 1.20 mol (Sb limiting) < 1.60 mol ($I_{2}$ limiting)       The limiting reactant is Sb. limiting reactant 1.20 mol < 1.60 mol       The theoretical yield is 1.20 mol Sb$I_{3}$ theoretical yield

 1.20 g Sb $\frac{1 mol Sb}{121.8 g Sb}$ × $\frac{2 mol SbI_{3}}{2 mol Sb}$ = 0.00985 mol Sb$I_{3}$ 2.40 g $I_{2}$ × $\frac{1 mol I_{2}}{253.8 g I_{2}}$ × $\frac{2 mol SbI_{3}}{3 mol I_{2}}$ = 0.006304 mol Sb$I_{3}$ 1. mol Sb$I_{3}$ 0.006304 mol ($I_{2}$ limiting) < 0.00985 mol (Sb limiting); thus $I_{2}$ is the limiting reactant. 0.006304 mol < 0.00985 mol       The theoretical yield is 0.006304 mol (3.17 g) Sb$I_{3}$. The reactant in excess is Sb. 2. limiting reactant theoretical yield 2.40 g $I_{2}$ × $\frac{1 mol I_{2}}{253.8 g I_{2}}$ × $\frac{2 mol Sb}{3 mol I_{2}}$ × $\frac{121.8 g Sb}{1 mol Sb}$ = 0.768 g Sb 3. mass Sb used up mass unreacted = mass present initially – mass used up = 1.20 g – 0.768 g = 0.43 g 4. mass unreacted