I have no idea how to solve these two equations. Can you please solve them and explain how you did it? my book is no help.
1. A strip of Magnesium is dropped into 6 M hydrochloric acid.
2. Chlorine gas is bubbled through a solution of potassium iodide.
2007-06-28 08:26:50 · 4 answers · asked by Marti 6 in Science & Mathematics ➔ Chemistry
Thanks, but also, what are these reactions known as? Combustion or decomp or what?
2007-06-28 08:37:27 · update #1
1. Mg + 2HCl ===> MgCl2 + H2
2. Cl2 + 2KI ===> 2KCl + I2
2007-06-28 08:31:03 · answer #1 · answered by steve_geo1 7 · 2⤊ 0⤋
They're both substitution reactions. In the first one, magnesium which is more electropositive, displaces hydrogen:
Mg + 2HCl -> MgCl2 + H2
On the second one, chlorine, which is more electronegative, displaces iodine:
2KI + Cl2 -> 2KCl + I2
2007-06-28 15:40:10 · answer #2 · answered by Israfel 3 · 0⤊ 1⤋
#1. Mg(s) + 2HCl(aq) >>> MgCl2(s) + H2(g)
#2. Cl2(g) + 2KI(s) >>> 2KCl(s) + I2(g)
Both of these reactions are considered single displacement reactions.
2007-06-28 20:44:23 · answer #3 · answered by scott k 4 · 0⤊ 0⤋
Mg + 2HCl--->MgCl2 +H2(g), hydrogen gas is produced
Cl2(g) +2 KI --->KCl +I2(s), Iodine is produced. If you collect it on a watch glass cooled with ice it is a blue black crystal. If you forget to cool it, it is a purple gas.
2007-06-28 15:34:48 · answer #4 · answered by science teacher 7 · 1⤊ 0⤋