Here\'s the combustion of ethane (not balanced):C2H6 + O2 ---> CO2 + H2O32.0 g o
ID: 76660 • Letter: H
Question
Here's the combustion of ethane (not balanced):C2H6 + O2 ---> CO2 + H2O32.0 g of ethane was burned with 15.0 grams of oxygen gas, and 10.8 grams of carbon dioxide was formed. Calculate the percent yield of carbon dioxide. % (number only, SF's, start with balanced equation and make sure you have enough O2 to burn all the ethane)Explanation / Answer
2C2H6 + 7O2 -> 4CO2 +6H2O Balance equation 32 g C2H6 * (1 molC2H6 / 30 C2H6 g) *(7mol O2/2mol C2H6) = 3.73 mols O2 required toreact with all C2H6 15 g O2 *(1mol O2/32 g O2) =0.46875 mol O2available So O2 is limiting ractant 15 g O2 * (1 mol O2/32 g) *(4 molCO2/7 mol O2) * (44 g CO2/1 molCO2) = 11.79 CO2 g (theoretical yield) % yield = actual yield/ theoretical yield = 10.8 g /11.79 g =91.6%