2253kJ/mol Calculation below Ca(2+) + 2 Cl | 242 | Ca(2+) + Cl2 | | | 348.7 * 2 | 1145 2 Cl(-) + Ca(2+) | | Ca(+) + Cl2 | | 590 | Ca(g) + Cl2(g) | 2253 | 178 | Ca(s) + Cl2(g) | | 795 | CaCl2(s)------------
It is:2Na(s) + Cl2(g) → 2NaCl(s)(But remember in exams you should know this kind of method) Peace out, add me on BBM its:12b6a7d9
The reason for this is that one molecule of Cl2 will have 2 atoms of Cl, joined together by a bond. This means that since 1 mole is 6.022 e23 times 2 atoms/molecule =1.204 e24 Cl atoms.
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Chlorine is produced at the anode. Brine at the cathodeOxidation reaction: 2 Cl- --'anode'--> Cl2 + 2e-
2253kJ/mol Calculation below Ca(2+) + 2 Cl | 242 | Ca(2+) + Cl2 | | | 348.7 * 2 | 1145 2 Cl(-) + Ca(2+) | | Ca(+) + Cl2 | | 590 | Ca(g) + Cl2(g) | 2253 | 178 | Ca(s) + Cl2(g) | | 795 | CaCl2(s)------------
-1, and since there is two, -2.
Cl2 everyone it is not that hard, 2 x chlorine is Cl2
It is:2Na(s) + Cl2(g) → 2NaCl(s)(But remember in exams you should know this kind of method) Peace out, add me on BBM its:12b6a7d9
13g Cl2 * (1mol cl / 34.45 ) * (2 mol / 2 mol) * (58.44g NaCl / 1 mol NaCl) = 22g NaCl
2Fe+3Cl2------>2FeCl3
Cl stands for chlorine or chloride as an periodic element
Chlorine gas is Cl2. Cl's atomic mass is 35.45 and that x 2 is 70.9
Ca + Cl2 ----> CaCl Step 1) Ca + Cl2 ----> 2 CaCl There are 2 Cl's on the reaction side so you need 2 on the Product side Step 2) 2Ca + Cl2 ----> 2 CaCl There are now 2 Ca on the product side and you need 2 on the reactant side Voila!
2NaCl + F2 -> 2NaF + Cl2 The first F in the equation has 2, so the second has to have 2 as well. But placing a 2 before the NaF, gives the Na 2. So add a 2 before the NaCl. and the Cl after the yield sign already has 2.
The reason for this is that one molecule of Cl2 will have 2 atoms of Cl, joined together by a bond. This means that since 1 mole is 6.022 e23 times 2 atoms/molecule =1.204 e24 Cl atoms.
Chlorine is produced at the anode. Brine at the cathodeOxidation reaction: 2 Cl- --'anode'--> Cl2 + 2e-