Butane, C_4 H_20, reacts with oxygen, O_2 to form water, H_2 O, and carbon dioxi
ID: 969753 • Letter: B
Question
Butane, C_4 H_20, reacts with oxygen, O_2 to form water, H_2 O, and carbon dioxide, CO_2, as shown in the following chemical equation: 2C_4H_10(g) + 18 O_2(g) rightarrow 10 H_2 O (g) + 8 CO_2(g) The coefficients in this equation represent mole ratios. Notice that the coefficient for water (10) is five times of that 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 dioxides (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 there ratios. Now that you knoe the molar masses of the relevant compounds, you are ready to start solving problems. in general the typical strategy is convert from grams of compound X to moles of compound X using the molar masses of compound X. Convert the moles of compound X of moles of compound Y using the coefficient in the balanced chemical equation. Convert the moles of compound Y to grams of compound Y using the molar mass of the compound Y. Calculate the mass of water produced when 5.06 g of butane reacts with excess oxygen. Express your answer to these significant figures and includes the appropriate units. Calculate the mass of butane needed to produces 53.1 g of carbon dioxide. Express your answer to these significant figures and include the appropriate units.Explanation / Answer
2C4H10(g) + 13 O2(g) ----> 10 H2O(g) + 8 CO2(g)
No of mol of butane = 5.05/58 = 0.087 mol
No of mol of H2O produced = 0.087*10/2 = 0.435 mol
mass of water produced = 0.435*18 = 7.83 grams
part C
No of mol of Co2 = 63.1/44 = 1.434 mol
No of mol of butane = 1.434*2/8 = 0.36 mol
mass of butane required = 0.36*58 = 20.88 grams
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