Question

Using the following equation, determine the % yield from the following reaction if 30.65 g of octane (C_8 8 ​ H_{18} 18 ​ ) react with excess oxygen to produce 81.75 g of CO_2 2 ​ (g) 2C_8 8 ​ H_{18} 18 ​ (I) + 25O_2 2 ​ (g) → 16CO_2 2 ​ (g) + 18H_2 2 ​ O(l)

Answer 1

86.4%

Step-by-step explanation:

`2C_(8)H_(18) + 25O_(2)->16CO_(2) + 18H_(2)O \\`

MM (C₈H₁₈) = 12*8+1*18 = 114

MM (CO₂) = 12 + 16*2 = 44

moles of C₈H₁₈ = 30.65 / 114 = 0.269

moles of CO₂ = 81.75 / 44 = 1.860

The % Yield is the amount of a product formed divided by the theorical amount formed if the limiting reactant reacts 100%.

The amount formed of CO₂ is 1.860 moles

The theorical amount formed is obtained with the rule of three and the stoichiometric coefficients.

2 moles of C₈H₁₈ produces 16 moles of CO₂

0.269 moles of C₈H₁₈ produces x moles of CO₂

x = 0.269 * 16 / 2 = 2.152 moles

`\%Yield = (Actual\ amount\ CO_(2))/(Theorical\ amount\ CO_(2))`

`\%Yield = (1.860)/(2.152) = 0.864 = 86.4%`

Answer 2

The yield of the reaction is 86.11%.

Step-by-step explanation:

2C₈H₁₈ (l) + 25O₂ (g) → 16CO₂ (g) + 18H₂O (l)

MW C₈H₁₈ = 114 g/mol

1 mol C₈H₁₈ _______ 114 g

x _______ 30.65 g

x = 0.27 mol C₈H₁₈

MW CO₂ = 44 g/mol

1 mol CO₂ _______ 44 g

x _______ 81.75 g

x = 1.86 mol CO₂

Considering a 100% yield reaction:

2 mol C₈H₁₈ ___________ 16 mol CO₂

0.27 mol C₈H₁₈ _________ y

y = 2.16 mol CO₂

2.16 mol CO₂ _______ 100%

1.86 mol CO₂ _______ z

z = 86.11%