Answer:
The yield of the reaction is 86.11%.
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%
Answer:
86.4%
Explanation:
[tex]2C_{8}H_{18} + 25O_{2}->16CO_{2} + 18H_{2}O \\[/tex]
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
[tex]\%Yield = \frac{Actual\ amount\ CO_{2}}{Theorical\ amount\ CO_{2}}[/tex]
[tex]\%Yield = \frac{1.860}{2.152} = 0.864 = 86.4%[/tex]
