STOICHIOMETRY/LIMITING REAGENTS CHM 1045LICHM 1046L POST LAB DATE SEQUENCE # A c

Question

STOICHIOMETRY/LIMITING REAGENTS CHM 1045LICHM 1046L POST LAB DATE SEQUENCE # A chemist mixes 3.02 grams of aluminum metal with 15.1 grams of elemental iodine. Write the formulas for: (a) A elemental aluminum elemental la | |-de umALI aluminunm odide odine 2) Write a balanced equation for the reaction. 3) How many moles of each reactant were mixed? Show calculation Moles elemental aluminunm 2 Moles 63.40 mph iodine 4) Demonstrate clearly which reactant is the limiting reagent. REA GET 5) How many grams of product could be formed? (Theoretical yield) Show calculation. 6) How many grams of excess reagent remained? II. Soluble barium salts are considered poisons and yet hospitals routinely administer barium sulfate "milkshakes" and enemas to highlight digestive tract X-rays. Explain this apparent contradiction

Explanation / Answer

1)

Elemental aluminum = Al

Elemental iodine = I2

Aluminum iodide= AlI3

2)

2 Al + 3 I2 = 2 AlI3

3) given that Al – 3.02 g

I2= 15.1 g

Number of moles = amount in g/ molar mass

Al moles = 3.02 g / 26.981 g/ mole

= 0.112 mole Al

Mole of I2: 15.1 g/ 253.80894 g/ mole

= 0.0595 mole I2

4)

2 Al + 3 I2 = 2 AlI3

I limiting agent has following properties:

0.0595 mole I2 * 2 Molel Al/ 3 mole I2= 0.04 mole Al

0.112 mole Al * 3 Mole I2/ 2 mole Al

= 0.168 mole I2

Thus I2 is limiting agent

5)

0.0595 mole I2 * 2 Molel AlI3 / 3 mole I2= 0.04 mole AlI3

Amount of AlI3 = number of moles * molar mass

= 0.04 mole AlI3 * 407.695 g/mol

= 16.31 g

6)

In this reaction Al is present in excess

Excess mole = total mole – used mole

= 0.112 mole Al- 0.04 mole Al

= 0.072 mole Al

Amount of Excess Al = 0.072 mole Al *26.981 g/ mole

= 1.94 g Al in excess

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.