Mass To Mass Relationships in Chemical Equations
Now we know that coefficients in a balanced chemical
equation can represent the number of particles, OR the
number of Moles.
So, we will now determine mass ratios for a reaction.
Remember the formula: m = M × n
mass of sample = molar mass × number of moles
Let’s look at the Haber Process (making Ammonia) again:
Balanced Equation:
Number of Particles:
Number of Moles:
Mass (g) *see below:
Total Mass:
N2 + 3H2 =
1
3
1
3
28.0 6.06
2NH3
2
2
34.06
Reactant
34.06 g
Product
34.06 g
*Remember Law of Conservation of Mass!
Mass N2 = 1 mole N2 × 28.0 g/mol N2 = 28.0 g
Mass H2 = 3 mole H2 × 2.02 g/mol H2 = 6.06 g
Mass NH3 = 2 mole NH3 × (14.0+3.03)g/mol = 34.06 g
Mass to Mass of Reactants Example:
In outer space, CO2 is produced by astronauts and must
be removed with LiOH to produce Lithium carbonate and
water. An astronaut, on average, produces 1.00 × 103g
CO2 per day. What mass of Lithium hydroxide is needed
per astronaut each day?
Answer:
Step 1: Write a Chemical reaction
CO2 + LiOH = Li2CO3 + H2O
Step 2: Balance your chemical reaction
CO2 + 2LiOH = Li2CO3 + H2O
Step 3: Given the number of grams of CO2, calculate the
number of moles of CO2 (we want to work with Mole
ratios!)
m=M×n
1.00 × 103 g = (12.00 g/mol + (2 × 16 g/mol)) × n
1.00 × 103 g = 44.00 g/mol × n
n = 22.7 mol CO2
Step 4: Use mole ratios to find # of moles of LiOH
Known ratio
= Unknown Ratio
2 mol LiOH
=
x mol LiOH
1 mol CO2
22.7 mol CO2
Cross Multiply and solve for x
(22.7 mol CO2) × 2 mol LiOH = x mol LiOH
1 mol CO2
45.4 mol LiOH = x mol LiOH
Step 5: convert number of moles of LiOH to grams
m=M×n
m = 45.4 mol LiOH × (6.90 + 16.00 + 1.01)* Look up #s
m = 45.4 mol LiOH × 23.91 g/mol LiOH
m = 1085.06 g
Therefore, 1.09 × 103 g of LiOH will react with 1.00 × 103
g CO2
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Summary:
1) Balanced Chem reaction
2) grams to moles (m = M × n)
3) mole : mole ratio (to find unknown)
4) moles to grams (m = M × n)
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Homework:
Page 244
Questions: 11, 12 and 13