For each of the reactions, calculate the mass (in grams) of the product formed when 15.59 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.
2K(s)+Cl2(g)−−−−−→2KCl(s)
Cl2 is underlinded
2K(s)+Br2(l)−−−−−→2KBr(s)
Br2 is underlined
4Cr(s)+3O2(g)−−−−−→2Cr2O3(s)
3O2 is unederlined
2Sr(s)−−−−+O2(g)→2SrO(s)
2Sr is underlined
thanks in advance
Chem help equations?
2017-10-06 12:52 pm
回答 (1)
2017-10-06 4:32 pm
(1)
Molar mass of Cl₂ = 35.45 × 2 g/mol = 70.90 g/mol
Molar mass of KCl = (39.01 + 35.45) g/mol = 74.46 g/mol
2K(s) + Cl₂(g) → 2KCl(s)
Mole ratio Cl₂ : KCl = 2 : 1
No. of moles of Cl₂ reacted = (15.59 g) / (70.90 g/mol) = 0.2199 mol
No. of moles of KCl produced = (0.2199 mol) × 2 = 0.4398 mol
Mass of KCl produced = (0.4398 mol) × (74.46 g/mol) = 32.75 g
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(2)
Molar mass of Br₂ = 79.90 × 2 g/mol = 159.80 g/mol
Molar mass of KBr = (39.01 + 79.90) g/mol = 118.91 g/mol
2K(s) + Br₂(l) → 2KBr(s)
Mole ratio Br₂ : KBr = 1 : 2
No. of moles of Br₂ reacted = (15.59 g) / (159.80 g/mol) = 0.09756 mol
No. of moles of KCl produced = (0.09756 mol) × 2 = 0.1951 mol
Mass of KCl produced = (0.1951 mol) × (118.91 g/mol) = 23.20 g
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(3)
Molar mass of O₂ = 16.00 × 2 g/mol = 32.00 g/mol
Molar mass of Cr₂O₃ = (52.00×2 + 16.00×3) g/mol = 152.00 g/mol
4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)
Mole ratio O₂ : Cr₂O₃ = 3 : 2
No. of moles of O₂ reacted = (15.59 g) / (32.00 g/mol) = 0.4872 mol
No. of moles of Cr₂O₃ produced = (0.4872 mol) × (2/3) = 0.3248 mol
Mass of Cr₂O₃ produced = (0.3248 mol) × (152.00 g/mol) = 49.37 g
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(4)
Molar mass of Sr = 87.62 g/mol
Molar mass of Sr = (87.62 + 16.00) g/mol = 103.62 g/mol
2Sr(s) + O2(g) → 2SrO(s)
Molar ratio Sr : SrO = 2 : 2 = 1 : 1
No. of moles Sr reacted = (15.59 g) / (87.62 g/mol) = 0.17793 mol
No. of moles of SrO produced = 0.17793 mol
Mass of SrO produced = (0.17793 mol) × (103.62 g/mol) = 18.44 g
Molar mass of Cl₂ = 35.45 × 2 g/mol = 70.90 g/mol
Molar mass of KCl = (39.01 + 35.45) g/mol = 74.46 g/mol
2K(s) + Cl₂(g) → 2KCl(s)
Mole ratio Cl₂ : KCl = 2 : 1
No. of moles of Cl₂ reacted = (15.59 g) / (70.90 g/mol) = 0.2199 mol
No. of moles of KCl produced = (0.2199 mol) × 2 = 0.4398 mol
Mass of KCl produced = (0.4398 mol) × (74.46 g/mol) = 32.75 g
====
(2)
Molar mass of Br₂ = 79.90 × 2 g/mol = 159.80 g/mol
Molar mass of KBr = (39.01 + 79.90) g/mol = 118.91 g/mol
2K(s) + Br₂(l) → 2KBr(s)
Mole ratio Br₂ : KBr = 1 : 2
No. of moles of Br₂ reacted = (15.59 g) / (159.80 g/mol) = 0.09756 mol
No. of moles of KCl produced = (0.09756 mol) × 2 = 0.1951 mol
Mass of KCl produced = (0.1951 mol) × (118.91 g/mol) = 23.20 g
====
(3)
Molar mass of O₂ = 16.00 × 2 g/mol = 32.00 g/mol
Molar mass of Cr₂O₃ = (52.00×2 + 16.00×3) g/mol = 152.00 g/mol
4Cr(s) + 3O₂(g) → 2Cr₂O₃(s)
Mole ratio O₂ : Cr₂O₃ = 3 : 2
No. of moles of O₂ reacted = (15.59 g) / (32.00 g/mol) = 0.4872 mol
No. of moles of Cr₂O₃ produced = (0.4872 mol) × (2/3) = 0.3248 mol
Mass of Cr₂O₃ produced = (0.3248 mol) × (152.00 g/mol) = 49.37 g
====
(4)
Molar mass of Sr = 87.62 g/mol
Molar mass of Sr = (87.62 + 16.00) g/mol = 103.62 g/mol
2Sr(s) + O2(g) → 2SrO(s)
Molar ratio Sr : SrO = 2 : 2 = 1 : 1
No. of moles Sr reacted = (15.59 g) / (87.62 g/mol) = 0.17793 mol
No. of moles of SrO produced = 0.17793 mol
Mass of SrO produced = (0.17793 mol) × (103.62 g/mol) = 18.44 g
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