If 23.5678 grams of octane (C_8H_18) liquid is reacted with 65.2212 grams of oxy

Question

If 23.5678 grams of octane (C_8H_18) liquid is reacted with 65.2212 grams of oxygen gas to produce carbon dioxide gas and water vapor, what is the limiting reactant, how many grams of water vapor is produced, how many grams of carbon dioxide gas is produced, and how many grams of the excess reagent will remain after the reaction ceases? What is the percent yield if in actuality 23.4528 grams of water vapor was produced? In the Hall-Heroult process for making aluminum, molten aluminum oxide is reacted with solid carbon to produce molten aluminum metal and carbon dioxide gas. If 67.8542 g of aluminum oxide and 11.2356 g of carbon are used, what is the limiting reactant, and how many grams of the excess reactant remain after the reaction ceased, and theoretically how many grams of aluminum could be produced? In reality the reaction produced a percent yield of 92.00%, how much aluminum was actually produced? 65.2148 g of carbon disulfide gas reacts with 132.7812 g of chlorine gas to produce carbon tetrachloride gas and disulfur dichloride gas. What is the limiting reactant, how many grams of the excess reactant remain after the reaction ceased, and theoretically how many grams of carbon tetrachloride could be produced? In reality the reaction produced 91.4572 grams of carbon tetrachloride gas, what was the percent yield?

Explanation / Answer

1) Data:

Molar Masses:

Octane: 114.23 g/mol

Oxygen (O2) : 32 g/mol

Water: 18 g/mol

CO2 = 44 g/mol

moles = mass / molar mass

moles of Octane = 23.5678 g/mol / 114.23 = 0.21 moles

moles of O2= 65.2212 g / 32 g/mol = 2.04 moles

Solution:

The balanced equation says that 2 moles of octane need 25 moles of O2 to react. The rate is 2:25. We have 0.21 moles of octane, which need 0.21 * (25/2) = 2.6 moles of oxygen and we only have 2.04. On the other hand, 2.04 moles of oxygen react with 2.04 * (2/25) = 0.1632 moles of octane. So, the limiting reactant is oxygen. In the end, we'll have 0.21 - 0.1632 = 0.0468 moles of octane of excess.

Theoritical Water produced:

(18 moles of water/ 25 moles of oxygen) * 2.04 moles of oxygen = 1.5 moles

mass of water = 1.5 moles * 18 g/mol = 27 g

Theoretical CO2 produced:

(16 moles of CO2 / 25 moles of oxygen) * 2.04 moles of oxygen = 1.31 moles

mass of CO2 = 1.31 moles * 44 g/mol = 57.64 g

Percent yield = (23.4528 / 27) *100% = 87%

2) Balanced eq.: 2 Al2O3 + 3 C ----> 3 CO2  + 4 Al (s)

Data:

Molar masses

Al2O3 : 101.96 g/mol

C : 12 g/mol

CO2 : 44 g/mol

Al: 27 g/mol

moles of Al2O3 = 67.8542 g / 101.96 g/mol = 0.665 moles

moles of C = 11.2356 g / 12 g/mol = 0.9363 moles

Solution:

2 moles of aluminium oxide need 3 moles of C. The rate is 2:3.

0.665 moles of aluminum oxide need:

0.665 * (3/2) = 0.9975 moles of C, but we only have 0.9363. That means Carbon is the limiting reactant.

0.9363 moles of Carbon need:

0.9363 * (2/3) = 0.6242 moles of A.O.

Excess of A.O = 0.665 - 0.6242 = 0.0408 moles

Theoretical Aluminium produced:

(4 moles of Al / 3 moles of C) * 0.9363 moles of C = 1.2484 moles

Theoretical Mass of Aluminium = 1.2484 moles * 27 g/mol = 33.71 g

If the real yield is 92%, that means that we have 0.92 of 33.71 g:

True Aluminium produced = 33.71 * 0.92 = 31 g

3) CS2 + 3 Cl2 ---> CCl4 + S2Cl2

Data:

Molar masses

CS2 : 76 g/mol

Cl2 : 70.9 g/mol

CCl4 :  153.8 g/mol

S2Cl2 : 134.9 g/mol

moles of CS2 = 65.2146 g / 76 g/mol = 0.858 moles

moles of Cl2 = 132.7812 g/ 70.9 g/mol = 1.873 moles

Solution:

1 mole of CS2 needs 3 moles of Cl2. The rate is 1:3. So, 0.858 moles of CS2 need 0.858*3 = 2.574 moles of Cl2 and we only have 1.873. That means Cl2 is the limiting reactant.

1.873 moles of Cl2 need:

1.873 *(1/3) = 0.624 moles of CS2

Excess of CS2 = 0.858 - 0.624 = 0.234 moles

Theoretical CCl4 produced:

1 moles CCl4 / 3 moles Cl2 * 1.873 moles Cl2 = 0.624 moles of CCl4

Mass of CCl4 = 0.624 moles * 153.8 g/mol = 96 g

Percent yield = (91.4572 / 96) * 100% = 95.3 %

Hope this helps!

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