Question is depicted below

Question

Gaseous butane CH3CH22CH3

 will react with gaseous oxygen O2

 to produce gaseous carbon dioxide CO2

 and gaseous water H2O

. Suppose 3.5 g of butane is mixed with 19.8 g of oxygen. Calculate the minimum mass of butane that could be left over by the chemical reaction. Round your answer to 2

 significant digits. 

Answer 1

2C4H10 + 13O2 ==> 8CO2 + 10H2O ... balanced equation for combustion of butanets,

Given the amounts of both reactants, we must first find the limiting reactant. One way to do this is to divide the MOLES of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For C4H10: 3.5 g x 1 mol / 58.1 g = 0.0602 mols (÷2->0.03)

For O2: 19.8 g x 1 mol / 32 g = 0.619 mols (÷13->0.048)

Since 0.03 is less than 0.048, C4H10 is our limiting reagent.

Because C4H10 (butane) is the limiting reactant, there should be none left over after the reaction.