How many moles is in 0.0688 Ag Cl
Ag +1, Cl -7
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
+1 for Ag, -1 for Cl
1 mole Cl = 35.453g Cl 28.4g Cl x 1mol Cl/35.453g Cl = 0.801 mole Cl
+1 for Ag -1 for Cl
0,5 moles Cl-
Ag +1, Cl -7
2 moles of NaCl, of course. Cl would definitely limit in this one to one reaction and you would have 19998 moles Na in excess.
it wouldn't be moles of Cl it would be Cl2 as chlorine doesn't exist as an atom it exsists as two joined to form a compound this is very easy stuff the answer is 15 moles of Cl2 as there 30 moles of Cl hope this helps learn your moles it's easy stuff
+1 for Ag, -1 for Cl
1 mole Cl = 35.453g Cl 28.4g Cl x 1mol Cl/35.453g Cl = 0.801 mole Cl
slightly
Bromide: Ag+ + Br- = AgBr which is a cream precipitate Chloride: Ag+ + Cl- = Ag Cl which is a white precipitate Iodide: Ag+ + I- = AgI which is a yellow precipitate
+1 for Ag -1 for Cl
slightly
Remember the equation moles = mass(g) / Mr Mr is the relative molecular mass. Refer to the Periodic Table 1 x Ag = 1 x 107.87 1 x Cl = 1 x 35.5 107.87 + 35.5 = 143.37 (Mr of AgCl) Substituting moles(AgCl) = 573.28 / 143.37 = 3.9986... ~ 4 moles (AgCl)
Based on the stoichiometry of NaCl, for every one mole of NaCl there is one mole of Na+ and one mole of Cl-. Therefore, there are 2.5 moles Na+ and 2.5 moles Cl-, totaling 5 moles of ions altogether.