Consider a reaction 2 C8H18 + 25 O 216 CO2 + 18 H2O
From reaction , 2 mole octane combines with 25 mole oxygen to give 16 moles of carbon dioxide.
Hence stoichiometric ratio of reactants is C8H18 : O 2 = 2 : 25 = 1 : 12.5
Provided molar ratio of reactants is C8H18 : O 2 = 2.43 : 4.86 = 1: 2
Comparing provided molar ratio with stoichiometric ratio of reactants it is clear that O 2 is limiting reactant , hence yield of product will depend upon amount of O 2.
From reaction we can write,
25 mole O 2 16 mole CO2
4.86 mole O 2 16 x 4.86 / 25 mole CO2
3.11 mole CO2
3.11 mole CO2 = 3.11 mol x ( 44.01 g / mol ) = 136.9 g CO2
Theoretical yield of product = 136.9 g
Practical yield of product = 39.92 g
% practical yield = [Practical yield of product / Theoretical yield of product ] x 100
= ( 39.92 g / 136.9 g ) x 100
= 29.2 %
give the percent yield when 39.92 g of Co2 are formed from the reaction of 2.43...
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Give the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 8.000 moles of O2.2 C8H18 + 25 O2 ? 16 CO2 + 18 H2Oa)12.50%b)25.00%c)20.00%d)50.00%
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2 C8H18 (l) + 25 O2 (g) ----> 16 CO2 (g) + 18 H2O (g)(OR 2 C8H18 (l) + 25 O2 (g) right arrow 16 CO2 (g) + 18 H2O (g))Use the following conversions: 1 mol = MW in g (2 d.p.) = 22.414 LAll gases are at STP. How many liters of CO2 would be produced from the reaction of 5.0 L of C8H18 with 55 L of O2?Theoretical yield of CO2 from 5.0 L C8H18 = _____________Theoretical yield...
When 10.58 g H2 react by the following balanced equation, 32.8 g H2O are formed. What is the percent yield of the reaction? 2 H2(g) + O2(g) → 2H2O(l) Select the correct answer below: 32.2% 34.7% 65.3% 38.9%
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