Question

CH 09 HW

± Masses of Reactants and Products

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± Masses of Reactants and Products 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.

Molar mass 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. Part A What is the molar mass of butane, C4H10? Express your answer to two decimal places and include the appropriate units. Hints The molar mass of C4H10 is based on the number of each kind of atom in the formula, and values from the periodic table: C4H10:    4(12.01 g/mole)+10(1.008 g/mole)=58.12 g/mole Similarly, the molar masses for H2O and CO2 are H2O:    2(1.008 g/mole)+(16.00 g/mole)=18.02 g/moleCO2:    (12.01 g/mole)+2(16.00 g/mole)=44.01 g/mole The molar mass can be used to convert between grams and moles of a compound. Solving stoichiometry problems that involve mass 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. Part B Calculate the mass of water produced when 7.35 g of butane reacts with excess oxygen. Express your answer to three digits and include the appropriate units. Hints Hint 1. Convert the mass of butane to moles The molar mass of butane, C4H10, is 58.12 g/mole . Convert 7.35 g of butane to moles. Express your answer to three digits and include the appropriate units. Hint 1. Identify how to convert from grams to moles(click to open) This hint will be shown after you complete previous hint(s). Hint 2. Part C Calculate the mass of butane needed to produce 68.3 g of carbon dioxide. Express your answer to three digits and include the appropriate units.

Answer 1

Given, the balanced chemical reaction,

2C₄H₁₀(g) + 13O₂(g) 10H₂O(g) + 8CO₂(g)

Part A)

Molar mass of C₄H₁₀ = [(4 x12.01) + (10 x 1.008)] = 58.12 g/mol

Part B) Given, the mass of butane = 7.35 g

Firstly calculating the number of moles of butane from the given mass and the molar mass of butane,

= 7.35 g of butane x ( 1 mol /58.12 g)

= 0.1265 mol of butane

Now, since oxygen is in excess, butane is the limiting reactant, thus using the moles of butane and the mole ratio from the balanced chemical reaction, calculating the number of moles of H₂O,

= 0.1265 mol of butane x ( 10 mol H₂O / 2 mol butane)

= 0.6323 mol of H₂O

Now, converting moles of H₂Oto grams,

= 0.6323 mol of H₂O x ( 18.02 g / 1 mol)

= 11.4 g of H₂O

Part C) Given, Mass of CO₂ = 68.3 g

Firstly calculating the number of moles of CO₂ by using the given mass and the molar mass of CO₂,

= 68.3 g CO₂ x ( 1 mol / 44.01 g)

= 1.5519 mol of CO₂

Now, using the moles of CO₂ and the mole ratio from the balanced chemical reaction, calculating the number of moles of butane,

= 1.5519 mol of CO₂ x ( 2 mol of butane / 8 mol of CO₂)

= 0.388 mol butane

converting moles of butane to grams by using the molar mass of butane,

= 0.388 mol butane x ( 58.12 g /1 mol)

= 22.5 g butane