You should have written CaCl2(aq) + 2AgNO3(aq) → Ca(NO3)2(aq) + 2AgCl(s) (The Precipitation reaction)
The full ionic equation though is ............................. Ca2+ + 2Cl- + 2Ag+ + 2NO3- → Ca2+ + 2NO3- + 2AgCl(s)
CaCl2 → Ca2+ and 2Cl-
ca + cl2 --> cacl2
silver nitrate (AgNO3) is an ionic compound and so does calcium chloride (CaCl2). Two ionic compounds react to produce two salts. This is called changing ions because Ag receives Cl now instead of NO3 and Ca receives NO3. Therefore, AgNO3 + CaCl2 = AgCl + Ca(NO3)2. A rule is present that Nitrate is soluble (it dissolves in H2O). Meanwhile, both, Cl and Ag are insoluble, therefore a solid we call as precipitate will settle to the bottom. this precipitate is AgCl (solid). All in all this reaction is called the precipitation reaction.
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
The balanced equation is :- Na2CO3 + CaCl2 = 2NaCl + CaCO3
2AgNO3 + CaCl2 ->2AgCl + Ca(NO3)2
2Ag(NO3)2+CaCl2 ---> 2AgCl +Ca(NO3)2
CaCl2 → Ca2+ and 2Cl-
Cu + AgNO3 --> Ag + Cu(NO3)2See formation of silver crystalshttp://www.youtube.com/watch?v=rgYhkVy5cBU
ca + cl2 --> cacl2
Calcium(Ca), being a more reactive element, reacts with FeCl3 to produce CaCl2 and Iron(Fe).3 Ca + 2 FeCl3 ----> 3 CaCl2 + 2Fe
silver nitrate (AgNO3) is an ionic compound and so does calcium chloride (CaCl2). Two ionic compounds react to produce two salts. This is called changing ions because Ag receives Cl now instead of NO3 and Ca receives NO3. Therefore, AgNO3 + CaCl2 = AgCl + Ca(NO3)2. A rule is present that Nitrate is soluble (it dissolves in H2O). Meanwhile, both, Cl and Ag are insoluble, therefore a solid we call as precipitate will settle to the bottom. this precipitate is AgCl (solid). All in all this reaction is called the precipitation reaction.
Ok, lets begin by writing out the reaction : 2AgNO3 +CaCl2 --> 2AgCl(s) + Ca(NO3)2 Precipitate = AgCl Now find the mol of compound in each solution: 14g AgNO3 x (mol/170g) = .082mol 4.83g CaCl2 x (mol/111g) = .044mol Determine limiting reactant: Notice in reaction that 2 CaCl2 molecules react with 1 AgNO3. Because 2(.044mol) > 1(.082mol), AgNO3 is your limiting reactant. Now that you know this you can find the mass of the precipitate .082molAgNO3x (2molAgCl/2molAgNO3)x(143.3g/molAgCl) = 11.75g b) Assuming all the AgNO3 is exhausted, there will be 2(.044)-(.082) = .006mol CaCl2 left .006mol x (111g/mol) = 0.67g CaCl2
The balanced equation is :- Na2CO3 + CaCl2 = 2NaCl + CaCO3
First it's CaCl2, with a lowercase L, not an i. The balanced equation is: Na2CO3(aq) + CaCl2(aq) --> 2NaCl(aq) + CaCO3(s)
2KCl + CaCO3
Ca + Cl2 -> CaCl2