ADVANCED HONORS CHEMISTRY
NAME:
REVIEW
DATE:
FORMULA AND EQUATION WRITING - ANSWERS PERIOD:
I. Directions: Write the chemical formulas for the following compounds. For all salts and acids, write oxidation
numbers before writing the final formula. Kudos to Mrs. Doyle for this worksheet.
1. cupric phosphate à Cu3(PO4)2
2. iron(III) carbonate à Fe2(CO3)3
3. plumbous chromate à PbCrO4
4. ferrous oxalate à FeC2O4
5. hypochlorous acid à HClO
6. magnesium hydroxide à Mg(OH)2
7. hydrobromic acid à HBr
8. nitric acid à HNO3
9. permanganic acid à HMnO4
10. sodium hexafluorosilicate à Na2SiF6
11. dichlorine heptoxide à Cl2O7
12. lithium sulfite à Li2SO3
13. ammonium selenate à (NH4)2SeO4
14. bismuth phosphide à BiP
15. chromium(III) nitrite à Cr(NO2)3
II. Directions: Write the correct name of the formula. In the case of a variable oxidation number, include the
names using both (Stock and Classical) systems.
16. CaCl2 • 6 H2O à calcium chloride hexahydrate
17. H3P à hydrophosphoric acid; hydrogen phosphide
18. SnS2 à tin(IV) sulfide; stannic sulfide
1
19. Cu2O à copper(I) oxide; cuprous oxide
20. ZnCO3 à zinc carbonate
21. AgCN à silver cyanide
22. SrS2O3 • 5 H2O à strontium thiosulfate pentahydrate
23. Al2(Cr2O7)3 à aluminum dichromate
24. H3AsO4 à arsenic acid
25. NH4HSO3 à ammonium bisulfite; ammonium hydrogen sulfite
26. K2B4O7 • 4 H2O à potassium tetraborate tetrahydrate
27. NiSe à nickel(II) selenide
28. Al2Te3 à aluminum telluride
29. CoBO3 à cobalt(III) borate; cobaltic borate
30. H3O+ à hydronium ion
III. Directions: Complete the following word equations by writing the predicted products. Then, write the
balanced chemical equation below the word equation.
31. sodium plus water à sodium hydroxide plus hydrogen
2 Na + 2 H2O à 2 NaOH + H2
32. calcium oxide plus hydrochloric acid à calcium chloride plus water
CaO + 2 HCl à CaCl2 + H2O
33. magnesium chloride plus bromine à NR
34. ferrous sulfide plus hydrochloric acid à ferrous chloride plus hydrosulfuric acid (hydrogen sulfide)
FeS + 2 HCl à FeCl2 + H2S
35. silver oxide à silver + oxygen
2 Ag2O à 4 Ag + O2
2 - AHC - Review - Formula and Equation Writing - Answers IV. Directions: Solve the following problem on lined paper, showing all work.
A plumber’s torch burns acetylene gas (C2H2) in the presence of oxygen. How many grams of water can be
produced by the reaction of 1.6 moles of acetylene with 4.25 moles of oxygen?
2 C2H2 (g) + 5 O2 (g) à 4 CO2 (g) + 2 H2O (g)
2 mol C2 H 2
1.6 mol C2 H 2
=
5 mol O 2
x
x = 4.0 moles O 2 needed
Since this plumber has 4.25 moles of oxygen gas available and
only needs 4.0 moles, there is an excess of oxygen gas.
Limiting Reactant = Acetylene gas
2 mol C2 H 2
1.6 mol C2 H 2
=
2 mol H 2O
x
x = 1.6 mol H 2O produced
Mass of H 2O produced = (moles of H 2O produced)(molar mass of H 2O)
! 1.6 mol H O $! 18.0 grams H O $
2
2
Mass of H 2O produced = #
&
&#
1
1
mol
H
O
"
%"
%
2
Mass of H 2O produced = 29 grams H 2O
3 - AHC - Review - Formula and Equation Writing - Answers