Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 32.2g of water is produced from the reaction of 42.24 of butane and of 62.3g oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.

Now that the reaction is balanced, we know that the water produced was 32.2 g. So, let's calculate the theorical yield of water using 42.24 g of butane and 62.3 g of oxygen.

The molar mass of butane is 58 g/mol, while water is 18 g/mol and oxygen 32 g/mol

The moles of each reactant are:

moles butane = 42.24 / 58 = 0.73 moles

moles oxygen = 62.3 / 32 = 1.95 moles

Now, let's calculate the limiting reactant:

1 mole butane -------> 13/2 mole oxygen

0.73 moles ---------> X

X = 0.73 * 13/2 = 4.75 moles oxygen

However, we only have 1.95 moles, therefore, oxygen is the limiting reactant.

As oxygen is the limiting reactant, this means that the moles produced of water will be:

moles water = 1.95 moles

Finally the mass of water:

m = 1.5 * 18 =¨35.1 g

The percent yield will be:

% = (32.2 / 35.1) * 100

% = 91.73 %