I have no idea what I'm doing can you explain step by step with these problems? Thanks for your help!
1. CaCO3 -> CaO + CO2
Determine how many grams of CaO can theoretically produced by 15.5 grams of CaCO3
2. C6H6O3 + 6O2 -> 6CO2 + 3 H2O
How any grams of H2O could be produced by 30.9 grams of C6H6O3 and excess oxygen?
Chemistry homework help?
2019-02-24 9:01 am
回答 (1)
2019-02-24 9:46 am
✔ 最佳答案
1.Molar mass of CaCO₃ = (40 + 12 + 16×3) g/mol = 100 g/mol
Molar mass of CaO = (40 + 16) g/mol = 56 g/mol
CaCO₃ → CaO + CO₂
1 mol CaCO₃ produces 1 mol CaO.
Moles of CaCO₃ reacted = (15.5 g) / (100 g/mol) = 0.155 mol
Moles of CaO produced = Moles of CaCO₃ reacted = 0.155 mol
Mass of CaO produced = (0.155 mol) × (56 g/mol) = 8.68 g
2.
Molar mass of C₆H₆O₃ = (12×6 + 1×6 + 16×3) g/mol = 126 g/mol
Molar mass of H₂O = (1×2 + 16) g/mol = 18 g/mol
C₆H₆O₃ + 6O₂ → 6CO₂ + 3H₂O
1 mol C₆H₆O₃ produces 3 mol H₂O.
Moles of C₆H₆O₃ reacted = (30.9 g) / (126 g/mol) = 0.245 mol
Moles of H₂O produced = (0.245 mol) × 3 = 0.735 mol
Mass of H₂O produced = (0.735 mol) × (18 g/mol) = 13.2 g
2019-02-24 9:07 am
(1)
15.5g CaCO3 --> 0.155 mol CaCO3
1 mol CaCO3 : 1 mol CaO
So, 0.155 mol CaO is produced --> 8.68g CaO produced.
[Use molar masses to go between moles and mass]
(2)
30.9g C6H6O3 --> 0.245 mol C6H6O3
1 mol C6H6O3 : 3 mol H2O
So, 0.735 mol H2O is produced --> 13.2 g H2O produced.
15.5g CaCO3 --> 0.155 mol CaCO3
1 mol CaCO3 : 1 mol CaO
So, 0.155 mol CaO is produced --> 8.68g CaO produced.
[Use molar masses to go between moles and mass]
(2)
30.9g C6H6O3 --> 0.245 mol C6H6O3
1 mol C6H6O3 : 3 mol H2O
So, 0.735 mol H2O is produced --> 13.2 g H2O produced.
收錄日期: 2021-05-01 22:30:37
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