Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Question

Gaseous butane CH3CH22CH3

 will react with gaseous oxygen O2

 to produce gaseous carbon dioxide CO2

 and gaseous water H2O

. Suppose 20. g of butane is mixed with 37.9 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits. 

Answer 1

This is a combustion reaction. The first thing to do when solving these problems is to setup the chemical equation and balance it.

CH₃CH₂CH₂CH₃ + (13/2) O₂ --> 4 CO₂ + 5 H₂O

From there, using the molar mass of each reactant, and the amount of starting material, determine the limiting reactant. For complete combustion, the fuel is completely consumed and oxygen is in excess.

Molar mass of butane: 58.12 g/mol starting mass of butane: 20. g starting moles: 0.34411 mol

Molar mass of oxygen: 32 g/mol starting mass of oxygen: 37.9 g starting moles: 1.18438 mol

Use dimensional analysis to determine how many moles of oxygen will be required to react all of the butane.

0.34411 mol butane * ((13/2 mol oxygen) / (1 mol butane)) = 2.2367 mol oxygen are required to react all of the butane.

I have 1.18438 mol of oxygen, not enough to react all of the butane, so oxygen is my limiting reactant and I can only make as much product as I have reactants.

I'll use dimensional analysis again to solve for the moles of water I can produce, and then use the molar mass of water (18.02 g/mol) to determine the maximum mass of water I can produce with the limiting reactant.

1.18438 mol oxygen * ((5 mol H₂O ) / ((13/2) mol oxygen)) = 0.9111 mol of water are produced.

0.9111 mol water * (18.02 g/mol) = 16.417 g water produced

The smallest number of sig figs from the values above is 2. So, our answer must also have 2 sig figs. This means we round down our answer to 16 g of water produced.