Chemistry, 08.10.2019 05:00 justin5647

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)

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Answer from: siniyalee30

86.4%

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

$\%Yield = \frac{Actual\ amount\ CO_{2}}{Theorical\ amount\ CO_{2}}$

$\%Yield = \frac{1.860}{2.152} = 0.864 = 86.4%$

Answer from: Quest

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Using the following equation, determine the % yield from the following reaction if 30.65 g of octane...

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