Stoichiometry worksheet

1. Mass of C₃H₆ used: 121.67 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{3C₃H₆ + 11O₃ → 9CO₃ + 6H₃O}$

2. Change grams of H₃O to moles by dividing by the weight of one mole (19.023 g/mol)

   $\frac{110\text{\hspace{0.17em}g}}{19.023\text{\hspace{0.17em}g/mol}}$$=5.782474\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 3 C₃H₆ for every 6 H₃O

   $5.782474\times \frac{3}{6}$$=2.891237\text{\hspace{0.17em}mol}$

4. Change moles of C₃H₆ back to grams by multiplying by the weight of one mole (42.081 g/mol)

   $2.891237\times 42.081$$=121.666141\text{\hspace{0.17em}g}$

2. Mass of H₃O produced: 107.85 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{3C₅H₁₁ + 14O₄ → 15CO₃ + 11H₃O}$

2. Change grams of C₅H₁₁ to moles by dividing by the weight of one mole (71.143 g/mol)

   $\frac{110\text{\hspace{0.17em}g}}{71.143\text{\hspace{0.17em}g/mol}}$$=1.546182\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 11 H₃O for every 3 C₅H₁₁

   $1.546182\times \frac{11}{3}$$=5.669333\text{\hspace{0.17em}mol}$

4. Change moles of H₃O back to grams by multiplying by the weight of one mole (19.023 g/mol)

   $5.669333\times 19.023$$=107.847715\text{\hspace{0.17em}g}$

3. Mass of CO₁ produced: 265.04 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{6C₆H₁₁ + 29O₂ → 36CO₁ + 22H₃O}$

2. Change grams of H₃O to moles by dividing by the weight of one mole (19.023 g/mol)

   $\frac{110\text{\hspace{0.17em}g}}{19.023\text{\hspace{0.17em}g/mol}}$$=5.782474\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 36 CO₁ for every 22 H₃O

   $5.782474\times \frac{36}{22}$$=9.46223\text{\hspace{0.17em}mol}$

4. Change moles of CO₁ back to grams by multiplying by the weight of one mole (28.01 g/mol)

   $9.46223\times 28.01$$=265.03706\text{\hspace{0.17em}g}$

4. Mass of O₁ used: 178.74 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{C₅H₉ + 19O₁ → 5CO₂ + 9H₁O}$

2. Change grams of H₁O to moles by dividing by the weight of one mole (17.007 g/mol)

   $\frac{90\text{\hspace{0.17em}g}}{17.007\text{\hspace{0.17em}g/mol}}$$=5.291939\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 19 O₁ for every 9 H₁O

   $5.291939\times \frac{19}{9}$$=11.17187\text{\hspace{0.17em}mol}$

4. Change moles of O₁ back to grams by multiplying by the weight of one mole (15.999 g/mol)

   $11.17187\times 15.999$$=178.738755\text{\hspace{0.17em}g}$

5. Mass of O₃ used: 437.99 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{C₄H₈ + 4O₃ → 4CO₂ + 4H₂O}$

2. Change grams of C₄H₈ to moles by dividing by the weight of one mole (56.108 g/mol)

   $\frac{128\text{\hspace{0.17em}g}}{56.108\text{\hspace{0.17em}g/mol}}$$=2.281315\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 4 O₃ for every 1 C₄H₈

   $2.281315\times \frac{4}{1}$$=9.125258\text{\hspace{0.17em}mol}$

4. Change moles of O₃ back to grams by multiplying by the weight of one mole (47.997 g/mol)

   $9.125258\times 47.997$$=437.985029\text{\hspace{0.17em}g}$

6. Mass of C₂H₇ used: 40.4 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{4C₂H₇ + 5O₃ → 8CO₁ + 7H₄O}$

2. Change grams of O₃ to moles by dividing by the weight of one mole (47.997 g/mol)

   $\frac{78\text{\hspace{0.17em}g}}{47.997\text{\hspace{0.17em}g/mol}}$$=1.625102\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 4 C₂H₇ for every 5 O₃

   $1.625102\times \frac{4}{5}$$=1.300081\text{\hspace{0.17em}mol}$

4. Change moles of C₂H₇ back to grams by multiplying by the weight of one mole (31.078 g/mol)

   $1.300081\times 31.078$$=40.403925\text{\hspace{0.17em}g}$

7. Mass of O₁ used: 439.29 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{C₃H₈ + 11O₁ → 3CO₃ + 2H₄O}$

2. Change grams of H₄O to moles by dividing by the weight of one mole (20.031 g/mol)

   $\frac{100\text{\hspace{0.17em}g}}{20.031\text{\hspace{0.17em}g/mol}}$$=4.992262\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 11 O₁ for every 2 H₄O

   $4.992262\times \frac{11}{2}$$=27.457441\text{\hspace{0.17em}mol}$

4. Change moles of O₁ back to grams by multiplying by the weight of one mole (15.999 g/mol)

   $27.457441\times 15.999$$=439.291598\text{\hspace{0.17em}g}$

8. Mass of CO₂ produced: 100 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{4C₂H₁₁ + 27O₁ → 8CO₂ + 11H₄O}$

2. Change grams of CO₂ to moles by dividing by the weight of one mole (44.009 g/mol)

   $\frac{100\text{\hspace{0.17em}g}}{44.009\text{\hspace{0.17em}g/mol}}$$=2.272262\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 8 CO₂ for every 8 CO₂

   $2.272262\times \frac{8}{8}$$=2.272262\text{\hspace{0.17em}mol}$

4. Change moles of CO₂ back to grams by multiplying by the weight of one mole (44.009 g/mol)

   $2.272262\times 44.009$$=100\text{\hspace{0.17em}g}$

9. Mass of C₁H₁₆ used: 9.28 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{C₁H₁₆ + 9O₂ → CO₂ + 16H₁O}$

2. Change grams of O₂ to moles by dividing by the weight of one mole (31.998 g/mol)

   $\frac{95\text{\hspace{0.17em}g}}{31.998\text{\hspace{0.17em}g/mol}}$$=2.968936\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 1 C₁H₁₆ for every 9 O₂

   $2.968936\times \frac{1}{9}$$=0.329882\text{\hspace{0.17em}mol}$

4. Change moles of C₁H₁₆ back to grams by multiplying by the weight of one mole (28.139 g/mol)

   $0.329882\times 28.139$$=9.282542\text{\hspace{0.17em}g}$

10. Mass of CO₄ produced: 618.8 g

1. Balance the equation. The numbers in front tell how many moles of each react together

   $\text{8C₂H₃ + 19O₄ → 16CO₄ + 12H₂O}$

2. Change grams of H₂O to moles by dividing by the weight of one mole (18.015 g/mol)

   $\frac{110\text{\hspace{0.17em}g}}{18.015\text{\hspace{0.17em}g/mol}}$$=6.106023\text{\hspace{0.17em}mol}$

3. Use the equation's ratio to change moles: 16 CO₄ for every 12 H₂O

   $6.106023\times \frac{16}{12}$$=8.141364\text{\hspace{0.17em}mol}$

4. Change moles of CO₄ back to grams by multiplying by the weight of one mole (76.007 g/mol)

   $8.141364\times 76.007$$=618.800629\text{\hspace{0.17em}g}$