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Chapter 3

Q. 3.10

Consider the reaction

2Sb(s) + 3I_{2}(s) → 2SbI_{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) + 3I_{2}(s) → 2SbI_{3}(s)]
Information given:
limiting reactant
theoretical yield
Asked for:

STRATEGY

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

ANALAYSIS
mass of each reactant: Sb (1.20 g), I_{2} (2.40 g)
balanced equation: [2Sb(s) + 3I_{2}(s) → 2SbI_{3}(s)]
Information given:
molar masses (MM) of SbI_{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 SbI_{3}.
2. The smaller number of moles SbI_{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

fig 3.8

Step-by-Step

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 SbI_{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 SbI_{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 SbI_{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 SbI_{3}

1. mol SbI_{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) SbI_{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