a) If 25.5g of calcium carbonate reacts with excess hydrogen phosphate, how many grams of carbon dioxide will be produced?
b) Calculate the number of grams of water that will be produced when 35.7 g of calcium carbonate is reacted with excess hydrogen phosphate.
c) What mass of hydrogen phosphate will react with excess calcium carbonate to produce 45.9g of calcium phosphate?
Solid calcium carbonate and aqueous hydrogen phosphate will yield aqueous calcium phosphate, gaseous carbon dioxide, and liquid water?
2016-03-02 4:14 pm
回答 (1)
2016-03-02 4:35 pm
a)
Molar mass of CaCO₃ = 40.1 + 12.0 + 16.0×3 = 100.1 g/mol
Molar mass of CO₂ = 12.0 + 16.0×2 = 44.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio CaCO₃ : CO₂ = 1 : 1
No. of moles of CaCO₃ = 25.5/100.1 mol
No. of moles of CO₂ produced = 22.5/100.1 mol
Mass of CO₂ produced = 44 × (22.5/100.1) = 9.89 g
b)
Molar mass of CaCO₃ = 100.1 g/mol
Molar mass of H₂O = 1.0×2 + 16.0 = 18.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio CaCO₃ : H₂O = 1 : 1
No. of moles of CaCO₃ = 35.7/100.1 mol
No. of moles of H₂O produced = 35.7/100.1 mol
Mass of H₂O produced = 18 × (35.7/100.1) = 6.42 g
c)
Molar mass of Ca₃(PO₄)₂ = 40.1×3 + 31.0×2 + 16.0×8 = 310.3 g/mol
Molar mass of H₃PO₄ = 1.0×3 + 31.0 + 16.0×4 = 98.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio H₃PO₄ : Ca₃(PO₄)₂ = 2 : 1
No. of moles of Ca₃(PO₄)₂ produced = 45.9/310.3 mol
No. of moles of H₃PO₄ needed = (45.9/310.3) × 2 mol
Mass of H₃PO₄ needed = 98.0 × (45.9/310.3) × 2 = 29.0 g
Molar mass of CaCO₃ = 40.1 + 12.0 + 16.0×3 = 100.1 g/mol
Molar mass of CO₂ = 12.0 + 16.0×2 = 44.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio CaCO₃ : CO₂ = 1 : 1
No. of moles of CaCO₃ = 25.5/100.1 mol
No. of moles of CO₂ produced = 22.5/100.1 mol
Mass of CO₂ produced = 44 × (22.5/100.1) = 9.89 g
b)
Molar mass of CaCO₃ = 100.1 g/mol
Molar mass of H₂O = 1.0×2 + 16.0 = 18.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio CaCO₃ : H₂O = 1 : 1
No. of moles of CaCO₃ = 35.7/100.1 mol
No. of moles of H₂O produced = 35.7/100.1 mol
Mass of H₂O produced = 18 × (35.7/100.1) = 6.42 g
c)
Molar mass of Ca₃(PO₄)₂ = 40.1×3 + 31.0×2 + 16.0×8 = 310.3 g/mol
Molar mass of H₃PO₄ = 1.0×3 + 31.0 + 16.0×4 = 98.0 g/mol
3 CaCO₃ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 3H₂O + 3CO₂
Mole ratio H₃PO₄ : Ca₃(PO₄)₂ = 2 : 1
No. of moles of Ca₃(PO₄)₂ produced = 45.9/310.3 mol
No. of moles of H₃PO₄ needed = (45.9/310.3) × 2 mol
Mass of H₃PO₄ needed = 98.0 × (45.9/310.3) × 2 = 29.0 g
收錄日期: 2021-04-18 14:35:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160302081437AAzVISB