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Solving for the products of reactions

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1. Fe²+ and Sb³-
2. Mg²+ and As³-
3. Al³+ and O²-
4. Co³+ and I-
5. Sr² + and P³-
6. Ca²+ and C4+
7. Fe³+ and As³-
8. Na + and S² ˉ

Write the name or formula for these compounds:
1. tin (II) acetate
2. bromous acid
3. copper(II) permanganate
4. CI4
5. SiBr4

Balance the following equations
a. Na2SO3 (aq) + S8 (s) → Na2S2O3
b. C4H11O 2 (l) + O 2 (g)→
c. C12H22O11 (l) + O 2 (g) →

Write the molecular equation for the following reactions ( show physical states of both reactants and products)
a. potassium hydroxide and copper (II) sulfate
b. ammonium sulfide and iron(III) chloride
c. barium nitrate and sodium carbonate
d. sodium chloride and silver nitrate

Write the net ionic equation for the following reactions show physical states for both reactants and products.
a. potassium carbonate and iron (III) nitrate
b. ammonium bromide and lead (IV) nitrate
c. sodium hydroxide and gold (III) chloride
d. lithium sulfide and mercury (I) nitrate

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Solution Preview


To do these questions, you need to make sure that the ionic compound formed is neutral. Therefore, the easy way to do this is to place the number next to the charge of one species as the subscript of the other species. For example, for the first one, we take the 2 next to the 2+ of Fe and put it as the subscript of Sb, making Sb2. Likewise; we take the 3 next to the 3- of Sb and put it as the subscript to Fe, making Fe3. Then put the two together making sure the cation (positively charged species) goes first. Do the same thing to each one:

1. Fe²+ and Sb³ ˉ  Fe3Sb2
2. Mg²+ and As³ ˉ  Mg3As2
3. Al³+ and O² ˉ  Al2O3
4. Co³+ and I ˉ  CoI3
5. Sr² + and P³ ˉ  Sr3P2
6. Ca²+ and C4+  Something is wrong here. You don't have any anion. Please check the question.
7. Fe³+ and As³ ˉ  FeAs
8. Na + and S² ˉ  Na2S

Write the name or formula for these compounds:
1. tin (II) acetate  Sn(C2H3O2)2
a. Each Sn (tin) cation has a +2 charge. Since acetate (C2H3O2) has a -1 charge, we need two of them to get the whole ...

Solution Summary

The solution provides many solutions to chemical reactions of different elements and compounds.