1.	26.0 gram Ca(ClO4)2	13.	5.08 gram XeF4		25.	10.0 gram KAl(SO4)2
2.	32.0 gram O2		14.	10.0 gram V2O5		26.	2.50 gram CoSO4 . 6H2O
3.	34.2 gram NH3		15.	2.50 gram K2Cr2O7	27.	24.0 gram CO
4.	9.00 gram H2SO4		16.	10.00 gram Na2CO3	28.	3.45 gram ZnCl2
5.	59.3 gram SnF2		17.	3.091 gram K2SO4	29.	36.0 gram Na2CrO2 . 4H2O
6.	0.00500 gram XeO3	18.	20.00 gram KOH		30.	15.0 gram PbO
7.	10.0 gram SO3		19.	0.0089 gram IF7		31.	50.00 gram KBr
8.	1.00 gram CO2		20.	32.58 gram CuS		32.	1.00 x 102 gram KCl
9.	5.00 gram CaCO3		21.	1.00 gram Ba(OH)2	33.	12.25 gram Sr(HCO3)2
10.	1.00 gram NaCl		22.	2.001 gram Al2O3	34.	0.00860 gram Ca3(PO4)2
11.	98.9 gram NaI		23.	2.00 x 10¯3 gram NH4NO3
12.	14.0 gram N2		24.	0.0010 gram Al(MnO4)3

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The Answers

Keep in mind that this is the technique involved:

These problems will give you the grams (for example, 26.0 g in the first problem) and ask you to calculate the moles. The moles will be your unknown and wind up in the left-hand denominator. The right side will be the molar mass in the numberator over 1.000 mol in the denominator. Solving this type of problem will ALWAYS end up with x being equal to the grams divided by the molar mass. For example, number three will be:

From this , we get:

34.2 g divided by 17.031 g mol¯¹ = 2.01 mol

1. 0.109

2. 1.00

3. 2.01

4. 0.0918

5. 0.378

6. 0.0000279

7. 0.125

8. 0.0227

9. 0.0500

10. 0.0171

11. 0.660

12. 0.500

13. 0.0245

14. 0.0550

15. 0.00850

16. 0.0945

17. 0.0177

18. 0.356

19. 0.0000342

20. 0.3408

21. 0.00584

22. 0.01962

23. 0.0000250

24. 0.0000026

25. 0.0387

26. 0.00950

27. 0.857

28. 0.0253

29.

30. 0.0672

31. 0.420

32. 1.34

33. 0.05843

34.