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.
Calculate the mass of water produced when 9.99 g of butane reacts with excess oxygen.
Express your answer to three significant figures and include the appropriate units.
Calculate the mass of butane needed to produce 70.1 g of carbon dioxide.
Express your answer to three significant figures and include the appropriate units.
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2017-10-06 10:35 pm
回答 (2)
2017-10-06 10:49 pm
✔ 最佳答案
Molar mass of C₄H₁₀ = (12.0×4 + 1.0×10) g/mol = 58.0 g/molMolar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
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(1) Calculate the mass of water produced when 9.99 g of butane reacts with excess oxygen.
2C₄H₁₀(g) + 13O₂(g) → 10H₂O(g) + 8CO₂(g)
Molar ratio C₄H₁₀ : H₂O = 2 : 10 = 1 : 5
No. of moles of C₄H₁₀ reacted = (9.99 g) / (58.0 g/mol) = 0.172 mol
No. of moles of H₂O produced = (0.1722 mol) × 5 = 0.860 mol
Mass of H₂O produced = (0.860 mol) × (18.0 g/mol) = 15.5 g
OR:
(9.99 g C₄H₁₀) × (1 mol C₄H₁₀ / 58.0 g C₄H₁₀) × (10 mol H₂O / 2 mol C₄H₁₀) × (18.01 g H₂O / 1 mol H₂O)
= 15.5 g H₂O
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(2) Calculate the mass of butane needed to produce 70.1 g of carbon dioxide.
2C₄H₁₀(g) + 13O₂(g) → 10H₂O(g) + 8CO₂(g)
Molar ratio C₄H₁₀ : CO₂ = 2 : 8 = 1 : 4
No. of moles of CO₂ produced = (70.1 g) / (44.0 g/mol) = 1.59 mol
No. of moles of C₄H₁₀ needed = (1.59 mol) × (1/4) = 0.398 mol
Mass of C₄H₁₀ needed = (0.398 mol) × (58.0 g/mol) = 23.1 g
OR:
(70.1 g CO₂) × (1 mol CO₂ / 44.0 g CO₂) × (2 mol C₄H₁₀ / 8 mol CO₂) × (58.0 g C₄H₁₀ / 1 mol C₄H₁₀)
= 23.1 g
2017-10-06 10:41 pm
For stoichiometry problems like this, the general process involves the following calculations:
mass of A --> moles of A --> moles of B ---> mass of B
Calculate the mass of water produced when 9.99 g of butane reacts with excess oxygen.
9.99 g C4H10 X (1 mol C4H10 / 58.12 g) X (10 mol H2O / 2 mol C4H10) X (18.01 g / 1 mol H2O) = 67.0 g H2O
Calculate the mass of butane needed to produce 70.1 g of carbon dioxide.
70.1 g CO2 X (1 mol / 44.00 g CO2) X (2 mol C4H10 / 8 mol CO2) X (58.12 g C4H10 / 1 mol) = 23.1 g C4H10
mass of A --> moles of A --> moles of B ---> mass of B
Calculate the mass of water produced when 9.99 g of butane reacts with excess oxygen.
9.99 g C4H10 X (1 mol C4H10 / 58.12 g) X (10 mol H2O / 2 mol C4H10) X (18.01 g / 1 mol H2O) = 67.0 g H2O
Calculate the mass of butane needed to produce 70.1 g of carbon dioxide.
70.1 g CO2 X (1 mol / 44.00 g CO2) X (2 mol C4H10 / 8 mol CO2) X (58.12 g C4H10 / 1 mol) = 23.1 g C4H10
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