What mass of silver chloride can be produced from 1.73 L of a 0.174 M solution of silver nitrate?
2016-10-03 1:29 pm
The reaction described in Part A required 3.56 L of calcium chloride. What is the concentration of this calcium chloride solution?
回答 (1)
2016-10-03 1:47 pm
CaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)
Mole ratio AgNO₃ : AgCl = 2 : 2 = 1 : 1
Number of moles of AgNO₃ = (0.174 mol/L) × (1.73 L)
Number of moles of AgCl = (0.174 mol/L) × (1.73 L)
Molar mass of AgCl = (107.9 + 35.5) = 143.4 g/mol
Mass of AgCl produced = (0.174 mol/L) × (1.73 L) × (143.4 g/mol) = 43.2 g
(to 3 sig. fig.)
Mole ratio CaCl₂ : Ag(NO₃)₂ = 1 : 2
Number of moles of AgNO₃ = (0.174 mol/L) × (1.73 L)
Number of moles of CaCl₂ = (0.174 mol/L) × (1.73 L) × (1/2)
Concentration of CaCl₂ = (0.174 mol/L) × (1.73 L) × (1/2) / (3.56 L) = 0.0423 M
(to 3 sig. fig.)
Mole ratio AgNO₃ : AgCl = 2 : 2 = 1 : 1
Number of moles of AgNO₃ = (0.174 mol/L) × (1.73 L)
Number of moles of AgCl = (0.174 mol/L) × (1.73 L)
Molar mass of AgCl = (107.9 + 35.5) = 143.4 g/mol
Mass of AgCl produced = (0.174 mol/L) × (1.73 L) × (143.4 g/mol) = 43.2 g
(to 3 sig. fig.)
Mole ratio CaCl₂ : Ag(NO₃)₂ = 1 : 2
Number of moles of AgNO₃ = (0.174 mol/L) × (1.73 L)
Number of moles of CaCl₂ = (0.174 mol/L) × (1.73 L) × (1/2)
Concentration of CaCl₂ = (0.174 mol/L) × (1.73 L) × (1/2) / (3.56 L) = 0.0423 M
(to 3 sig. fig.)
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