if 2.00 grams of co2 were produced when 4.00 grams of c5h12 were combusted in excessoxygen, what is the percent yield of the combustion reaction?

The % yield of the reaction is 16.4 %

Explanation:

Step 1: Data given

Mass of CO2 =2.00 grams

Mass of C5H12 = 4.00 grams

O2 is in excess

Molar mass of CO2 = 44.01 g/mol

Molar mass of C5H12 = 72.15 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

C5H12 + 8O2 → 5CO2 + 6H2O  

Step 3: Calculate moles C5H12

Moles C5H12 = mass C5H12 / molar mass C5H12

Moles C5H12 = 4.00 grams / 72.15 g/mol

Moles C5H12 = 0.0554 moles

Step 4: Calculate moles of CO2

For 1 mol pentane, we need 8 moles of O2 to produce 5 moles CO2 and 6 moles H2O

For 0.0554 moles C5H12 we'll have 5*0.0554 = 0.277 moles CO2

Step 5: Calculate mass CO2

Mass CO2 = moles CO2 * molar mass CO2

Mass CO2 = 0.277 moles ¨44.01 g/mol

Mass CO2 = 12.2  grams CO2

Step 6: Calculate % yield

% yield = (actual yield / theoretical yield) *100%

% yield = (2.00 grams / 12.2 grams) * 100%

% yield = 16.4 %

The % yield of the reaction is 16.4 %

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.