Percent yield?

Question

Gaseous butane (CH3(CH2)2CH3) reacts with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 0.256 g of water is produced from the reaction of 1.2 g of butane and 1.6 g of oxygen, calculate the percent yield of water.                                                                                                                                                                       So far i have balanced the equation to be 2 (CH3(CH2)2)CH3) + 13 O2————> 8 CO2 + 10 H2O

 

Answer 1

Balancing the reaction is a great start!  It gives you the mole ratios of each reactant and product.  Now you need to calculate the expected amount of water from each reactant.

 

1.2 g butane x 1 mol butane/58 g butane x 10 mol water/2 mol butane x 18 g water/mol water = 1.86 g water

 

1.6 g oxygen x 1 mol oxygen/32 g oxygen x 10 mol water/13 mol oxygen x 18 g water/mol water = 0.692 g water

 

You will run out of oxygen before you run out of butane, so oxygen is the limiting reactant and you would expect to produce 0.692 g of water.  Since you only produced 0.256 g, you then use that to calculate the percentage yield.

 

% yield = (0.256/0.692) x 100 = 36.99%

 

Using two significant figures gives 37% yield.