Moles & Grams, Stoichiometry Worksheet – Chemistry
The key terms in this Stoichiometry worksheet include Moles, Grams.
SHOW WORK on separate paper and copy final answers in the boxes provided. Use correct number of significant figures and report units with answers where required.
- Find the number of moles in 25.0 grams of each of the following:
Ni b. C6H6 c. Ca(OH)2
Calculations
Find the number of moles in 25.0 grams of each of the following:
Ni
25.0 g x 1/58.7 = 0.43 mol
C6H6
25.0 g x 1/78.1 = 0.32 mol
Ca(OH)2
25.0 g x 1/74.09 = 0.34 mol
- Find the mass of 0.750 moles of each of the following:
Br2 b. P4O10 c. Cu2S
Calculations
Find the mass of 0.750 moles of each of the following:
Br2
0.750 x 159.8 = 119.86 g
P4O10
0.750 x 283.89 = 212.92 g
Cu2S
0.750 x 159.7 = 119.8 g
- Find the number of particles in 35.0 grams of each of the following:
- Ag b. CF4 c. C6H12O6
Calculations
Find the number of particles in 35.0 grams of each of the following:
Ag
35 x 1/107.9 = 0.324 mol
1 mol = 6.022 x 1023 particles
0.324 x 6.022 x 1023 = 1.95 x 1023 particles
CF4
35 x 1/88 = 0.398 mol
1 mol = 6.022 x 1023 particles
0.398 x 6.022 x 1023 = 2.395 x 1023 particles
C6H12O6
35 x 1/180.2 = 0.194 mol
1 mol = 6.022 x 1023 particles 0.194 x 6.022 x 1023 = 1.169 x 1023 particles
SHOW WORK on separate paper and copy final answers in the boxes provided. Use correct number of significant figures and report units with answers where required.
- Find the number of moles in 25.0 grams of each of the following:
Ni b. C6H6 c. Ca(OH)2
Calculations
Find the number of moles in 25.0 grams of each of the following:
Ni
25.0 g x 1/58.7 = 0.43 mol
C6H6
25.0 g x 1/78.1 = 0.32 mol
Ca(OH)2
25.0 g x 1/74.09 = 0.34 mol
- Find the mass of 0.750 moles of each of the following:
Br2 b. P4O10 c. Cu2S
Calculations
Find the mass of 0.750 moles of each of the following:
Br2
0.750 x 159.8 = 119.86 g
P4O10
0.750 x 283.89 = 212.92 g
Cu2S
0.750 x 159.7 = 119.8 g
- Find the number of particles in 35.0 grams of each of the following:
- Ag b. CF4 c. C6H12O6
Calculations
Find the number of particles in 35.0 grams of each of the following:
Ag
35 x 1/107.9 = 0.324 mol
1 mol = 6.022 x 1023 particles
0.324 x 6.022 x 1023 = 1.95 x 1023 particles
CF4
35 x 1/88 = 0.398 mol
1 mol = 6.022 x 1023 particles
0.398 x 6.022 x 1023 = 2.395 x 1023 particles
C6H12O6
35 x 1/180.2 = 0.194 mol
1 mol = 6.022 x 1023 particles 0.194 x 6.022 x 1023 = 1.169 x 1023 particles