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1) HCl + Na4Cl =
2)HCl + KIO3 =
3)H2SO4 + HNO3 =
4)H2SO4 + Na2SO3 =
5)H2SO4 + NH4Cl =
6)H2SO4 + KIO3 =
7)H2SO4 + BaCl2 =
8)Na2SO3 + HNO3 =
9)NH4Cl + HNO3 =
10)KIO3 + HNO3=
11)BaCl2 + HNO3 =
12)Na2SO3 + NH4Cl =
13)Na2SO3 + KIO3 =
14)Na2SO3 + BaCl2=
15)KIO3 + BaCl2=
Any of these are greatly appreciated... Thank You.

2007-12-11 14:08:36 · 2 answers · asked by Anonymous in Science & Mathematics Chemistry

2 answers

The way to do these is first finish the reaction with the possible products that could be formed: (usually you don't need to balance the equation as long as you know the possible products).

The other thing you need is: somewhere in your text, there should be a table of solubilities of inorganic compounds in water. Usually, it will have all the common + chaged ions on one axis and all the common - charged ions on the other and where they intersect there will be designations "s" soluble, "i" insoluble, "ss" slightly soluble etc. (there will be a few combinations that don't exist)

I don't have time to do all of your homework: I'll do a couple so you can see how it works:

#7 H2SO4 + BaCl2 ---> BaSO4 + 2 HCl

The possible products are BaSO4 and HCl. We know HCl is soluble, it is hydrochloric acid (but you could look up the properties of the gas HCl if necessary, the other product; BaSO4 you should be able to look up in the table I described above and you will find that it is insoluble and there will be a precipitate formed. (OK, if you want to get really anal, a tensy weensy nano-smidgen dissolves, but it is essentially insoluble)

Here's another:

#9 NH4Cl + HNO3 ---> NH4NO3 + HCl

In this case, both of the possible products are soluble, so no precipitate will form in this case.

2007-12-11 14:43:17 · answer #1 · answered by Flying Dragon 7 · 0 1

http://www.kentchemistry.com/links/Kinetics/PredictingDR.htm

2007-12-11 22:42:43 · answer #2 · answered by kentchemistry.com 7 · 0 1

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