Question

Butane, C4H10, reacts with oxygen, O2, to form water, H2O, and carbon dioxide, CO2, as shown in the following chemical equation: 2C4h10(g)+13O2(g)->10H2O(g)+8CO2(g). Calculate the mass of water produced when 1.77 grams of butane reacts with excessive oxygen? Calculate the mass of butane needed to produce 71.6 of carbon dioxide

Answer 1

From the given reaction in 1.77 grams of butane, the mass of water produced will use the 10 mol H2O: 2 C4H10 stoichiometric ratio. Calculating, the amount of water produced is 2.75 grams water. The mass of butane needed from the same reaction nd 71.6 g CO2 is 23.60 g butane.