COMBINING STOICHIOMETRY AND MOLAR MASS CONVERSIONS (MOLES TO GRAMS CONVERSIONS):
Now that we are more comfortable with stoichiometry, we can combine it with our previous efforts of molar mass conversions in the how to convert between grams and moles section. This means we are going to further complicate our conversion map to include more destinations.
VIDEO Stoichiometry Conversions Demonstrated Example 3: If you have 4 mol of O_{2} then how many grams of H_{2} will you need to completely react? You will need a periodic table to help solve this problem.
2 H_{2(g)} + O_{2(g)} —-> 2 H_{2}O_{(l)}
Step 1:
What information does the question supply us with?
Answer: 4 mol O_{2}
Step 2:
What units does the question ask?
Answer: g H_{2}
Step 3:
How many conversions must we do?
Answer: Look at the conversion map. We pass through 2 arrows when we go from Moles of A —> Moles of B —> Grams of B. 2 arrows = 2 conversions
Step 4:
How do we set up the problem?
Answer:
4 mol O_{2} | g H_{2} | ||
1 |
Step 5:
What is the first conversion?
Answer: mole to mole ratio (coefficient ratio)
Step 6:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
4 mol O_{2} | H_{2} | g H_{2} | |
O_{2} |
Step 7:
What is the next step?
Answer: Fill in the numbers and cross out units
4 mol O_{2} | 2 H_{2} | g H_{2} | |
1 O_{2} |
Step 8: Simplify
4 mol | 2 H_{2} | g H_{2} | |
1 |
Step 9:
What is the next conversion?
Answer: molar mass (grams to mole ratio) of H_{2} found on the periodic table
Step 10:
How do I set it up?
Answer: units first, set up the units that you need to cancel out (in red)
4 mol | 2 H_{2} | g = | g H_{2} |
1 | mol |
Step 11:
What is the next step?
Answer: Fill in numbers and cross out units
4 mol | 2 H_{2} | 2 g = | g H_{2} |
1 | 1 mol |
Step 12: Simplify
4 | 2 H_{2} | 2 g = | g H_{2} |
1 | 1 |
Step 13:
How do I know I am done with conversions?
Answer: The only units left are the units that match the answer. In this case g and H_{2}
4 | 2 H_{2} | 2 g = | g H_{2} |
1 | 1 |
Step 14:
How do I do the calculations?
Answer: (4 * 2 * 2) = 16
4 | 2 H_{2} | 2 g = | 16 g H_{2} |
1 | 1 |
Step 15:
COMPLETE ANSWER: 16 g H_{2}
VIDEO Stoichiometry Conversions Demonstrated Example 4: If you have 30g of CO_{2} then how many moles of C_{2}H_{6} will be produced? You will need a periodic table to help solve this problem.
2 C_{2}H_{6(l)} + 7 O_{2(g)} ——> 4 CO_{2(g)} + 6 H_{2}O_{(g)}
Step 1:
What information does the question supply us with?
Answer: 30g CO_{2}
Step 2:
What units does the question ask?
Answer: mol C_{2}H_{6 }
Step 3:
How many conversions must we do?
Answer: Look at the conversion map. We pass through 2 arrows when we go from Grams of A —> Moles of A —> Moles of B. 2 arrows = 2 conversions
Step 4:
How do we set up the problem?
Answer:
30 g CO_{2} | mol C_{2}H_{6 } | ||
1 |
Step 5:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of CO_{2} found on the periodic table
Step 6:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
30 g CO_{2} | mol | mol C_{2}H_{6 } | |
g |
Step 7:
What is the next step?
Answer: Fill in the numbers and cross out units
30 g CO_{2} | 1 mol | mol C_{2}H_{6 } | |
44 g |
Step 8: Simplify
30 CO_{2} | 1 mol | mol C_{2}H_{6 } | |
44 |
Step 9:
What is the next conversion?
Answer: mole to mole ratio (coefficient ratio)
Step 10:
How do I set it up?
Answer: units first, set up the units that you need to cancel out (in red)
30 CO_{2} | 1 mol | C_{2}H_{6 } | mol C_{2}H_{6 } |
44 | CO_{2} |
Step 11:
What is the next step?
Answer: Fill in numbers and cross out units
30 CO_{2} | 1 mol | 2 C_{2}H_{6 } | mol C_{2}H_{6 } |
44 | 4 CO_{2} |
Step 12: Simplify
30 | 1 mol | 2 C_{2}H_{6 } | mol C_{2}H_{6 } |
44 | 4 |
Step 13:
How do I know I am done with conversions?
Answer: The only units left are the units that match the answer. In this case mol and C_{2}H_{6 }
30 | 1 mol | 2 C_{2}H_{6} = | mol C_{2}H_{6 } |
44 | 4 |
Step 14:
How do I do the calculations?
Answer: (30 * 2) / (44 * 4) = 0.34
30 | 1 mol | 2 C_{2}H_{6} = | 0.34 mol C_{2}H_{6 } |
44 | 4 |
Step 15:
COMPLETE ANSWER: 0.34 mol C_{2}H_{6 }
VIDEO Stoichiometry Conversions Demonstrated Example 5: If you have 12.9 g of Ca_{3}(PO_{4})_{2 } then how many grams of NaCl will you need to completely react? You will need a periodic table to help solve this problem. You can also use the conversion map.
2 Na_{3}PO_{4(aq)} + 3 CaCl_{2(aq)} ——> Ca_{3}(PO_{4})_{2(s)} + 6 NaCl_{(aq)}
Step 1:
What information does the question supply us with?
Answer: 12.9 g Ca_{3}(PO_{4})_{2 }
Step 2:
What units does the question ask?
Answer: g NaCl
Step 3:
How many conversions must we do?
Answer: Look at the conversion map. We pass through 3 arrows when we go from Grams of A —> Moles of A —> Moles of B —> Grams of B. 3 arrows = 3 conversions
Step 4:
How do we set up the problem?
Answer:
12.9 g Ca_{3}(PO_{4})_{2 } | g NaCl | |||
1 |
Step 5:
What is the first conversion?
Answer: molar mass (grams to mole ratio) of Ca_{3}(PO_{4})_{2} found on the periodic table
Step 6:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
12.9 g Ca_{3}(PO_{4})_{2 } | mol | g NaCl | ||
g |
Step 7:
What is the next step?
Answer: Fill in the numbers and cross out units
12.9 g Ca_{3}(PO_{4})_{2 } | 1 mol | g NaCl | ||
310 g |
Step 8: Simplify
12.9 Ca_{3}(PO_{4})_{2 } | 1 mol | g NaCl | ||
310 |
Step 9:
What is the next conversion?
Answer: mole to mole ratio (coefficient ratio)
Step 10:
How do I set it up?
Answer: units first, set up the units that you need to cancel out (in red)
12.9 Ca_{3}(PO_{4})_{2} | 1 mol | NaCl | g NaCl | |
310 | Ca_{3}(PO_{4})_{2} |
Step 11:
What is the next step?
Answer: Fill in the numbers and cross out units
12.9 Ca_{3}(PO_{4})_{2} | 1 mol | 6 NaCl | g NaCl | |
310 | 1 Ca_{3}(PO_{4})_{2} |
Step 12: Simplify
12.9 | 1 mol | 6 NaCl | g NaCl | |
310 | 1 |
Step 13:
What is the next conversion?
Answer: molar mass (grams to mole ratio) of NaCl found on the periodic table
Step 14:
How do I put that in?
Answer: units first, set up the units that need to cancel out (in red)
12.9 | 1 mol | 6 NaCl | g = | g NaCl |
310 | 1 | mol |
Step 15:
What is the next step?
Answer: Fill in the numbers and cross out units
12.9 | 1 mol | 6 NaCl | 58 g = | g NaCl |
310 | 1 | 1 mol |
Step 16: Simplify
12.9 | 1 | 6 NaCl | 58 g = | g NaCl |
310 | 1 | 1 |
Step 17:
How do I know I am done with conversions?
Answer: The only units left are the units that match the answer. In this case g and NaCl
12.9 | 1 | 6 NaCl | 58 g = | g NaCl |
310 | 1 | 1 |
Step 18:
How do I do the calculations?
Answer: (12.9 * 6 * 58) / (310) = 14.5
12.9 | 1 | 6 NaCl | 58 g = | 14.5 g NaCl |
310 | 1 | 1 |
Step 19:
COMPLETE ANSWER: 14.5 g NaCl
PRACTICE PROBLEMS: Calculate the moles or grams you can obtain from the moles or grams you are given. You will need the periodic table.
If you have 3 moles of Mg then how many grams of O_{2} will you need?
2 Mg + O_{2} —> 2 MgO
Answer: 48 g O_{2}
If 25 g of Ca(OH)_{2} are used, how many moles of NaOH can you make?
Ca(OH)_{2} + Na_{2}S —> CaS + 2 NaOH
Answer: 0.68 mol NaOH
How many grams of Al do you need if you have 8 g of Br_{2}?
2 AlBr_{3} —> 2 Al + 3 Br_{2}
Answer: 0.9 g Al
If you have 0.6g of CO_{2}, how many grams of O_{2} can you make?
2 C_{4}H_{10 } + 13 O_{2} —> 8 CO_{2} + 10 H_{2}O
Answer: 0.71 g O_{2}