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) rightarrow 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. The first step of many stoichiometry problems is to convert the given value from grams to moles. Molar masses, which can be found using the periodic table, serve as conversion factors between grams and moles. What is the molar mass of butane. C4H10? Express your answer numerically in grams per mole using four significant figures. Now that you know the molar masses of the relevant compounds, you are ready to start solving stoichiometry problems. In general, the typical strategy is Convert from grams of compound X to moles of compound X using the molar mass of compound X. Convert from moles of compound X to moles of compound Y using the coefficients in the balanced chemical equation. Convert from moles of compound Y to grams of compound Y using the molar mass of compound Y. Calculate the mass of water produced when 7.81g of butane reacts with excess oxygen. Express the mass of water numerically in grams. Calculate the mass of water produced when 7.81g of butane reacts with excess oxygen Express the mass of water numerically in grams. Calculate the mass of butane needed to produce 64.6g of carbon dioxide.
Explanation / Answer
Part B. molecular mass of butane = 4 x 12 + 1 x 10 = 58.12 g
mol of butane = 7.81/58.12 = 0.134 mol
mol of water = 10 x 0.134 / 2 = 0.672 mol
mass = (16 + 2) x 0.672 = 12.10 g
Part c. 8 x (12 + 32) x= 352g CO2 production will require 58.12 x 2 g butane
so 64.6g will = 58.12 x 2 x 64.6 / 352 = 21.33 g
