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) The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times that of butane (2). Thus, the number of moles of water produced is five times the number of moles of butane that react. Also, notice that the coefficient for butane (2) is one-fourth the coefficient of carbon dioxide (8). Thus, the number of moles of butane that react is one-fourth the number of moles of carbon dioxide that you produce. But be careful! If you are given the mass of a compound, you must first convert to moles before applying these ratios. A) Calculate the mass of water produced when 3.35 g of butane reacts with excess oxygen. B) Calculate the mass of butane needed to produce 66.7 g of carbon dioxide.

Answer 1

A)

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

From the balanced chemical equation ratio of moles of butane to the moles of H2O produced will always be 2/10 = 1/5 So, for 1 mole of butane 5 moles of H2O will be produced. So, 3.35 g of butane = 3.35/58.1 moles of butane (Molar mass of butane = 58.1) = 0.057 moles of butane.

Now, the number of moles of H2O produced will be 5*0.057 moles = 0.288 moles

Mass of 0.288 moles of H2O = 0.288*18 = 5.18 gm

B)

Ratio of moles of butane to the moles of CO2 produced will always be 2/8 = 1/4 .

66.7 g CO2 = 66.7/44 moles of CO2 gas (Molar mass of CO2 = 44) = 1.51 moles of CO2

If for 4 moles of CO2 gas we need 1 mole of butane gas. For 1.51 moles of CO2 gas we will need 1.51/4 moles of butane which is equal to 0.377 moles.

0.377 moles of butane = 58.12*0.377 = 21.91 gm of butane