To find the mass of water (H₂O) produced from the metabolism of 1.2 kg of tristearin (C₅₇H₁₁₀O₆), we first need to write the chemical equation for the complete combustion of tristearin. The general reaction is:
\[ \text{C}{57}\text{H}{110}\text{O}_6 + O_2 \rightarrow CO_2 + H_2O \]
We will first balance this combustion reaction to understand how many moles of H₂O are produced from one mole of tristearin.
Step 1: Balancing the Reaction
The unbalanced reaction looks like this:
\[ \text{C}{57}\text{H}{110}\text{O}_6 + O_2 \rightarrow CO_2 + H_2O \]
To balance carbon, hydrogen, and oxygen:
- Carbon (C): There are 57 carbon atoms, so we need 57 CO₂.
- Hydrogen (H): There are 110 hydrogen atoms, which means we need 55 H₂O (since each H₂O has 2 hydrogen atoms).
- Oxygen (O): We must account for the oxygen from both CO₂ and H₂O and the oxygen in tristearin.
Calculating the oxygen:
- From 57 CO₂: 57 × 2 = 114 O
- From 55 H₂O: 55 × 1 = 55 O
- Total O needed = 114 + 55 = 169 O
Now, consider the oxygen in the tristearin molecule:
- Tristearin contains 6 O, so we need to subtract this from the total oxygen needed from O₂:
- Required O from O₂ = 169 - 6 = 163 O
Since each O₂ molecule contains 2 O: \[ \text{O}_2 \text{ required} = \frac{163}{2} = 81.5 \text{ O}_2 \]
Thus, the balanced reaction is: \[ \text{C}{57}\text{H}{110}\text{O}_6 + 81.5 \text{ O}_2 \rightarrow 57 \text{ CO}_2 + 55 \text{ H}_2O \]
Step 2: Moles of Tristearin and H₂O
Next, we calculate the molar mass of tristearin (C₅₇H₁₁₀O₆):
- C: 12.01 g/mol
- H: 1.008 g/mol
- O: 16.00 g/mol
Calculating the molar mass: \[ \text{Molar mass of C}{57}\text{H}{110}\text{O}_6 = (57 \times 12.01) + (110 \times 1.008) + (6 \times 16.00) \]
Calculating each component:
- C: \( 57 \times 12.01 = 685.57 \) g
- H: \( 110 \times 1.008 = 110.88 \) g
- O: \( 6 \times 16.00 = 96.00 \) g
Total molar mass: \[ 685.57 + 110.88 + 96.00 = 892.45 \text{ g/mol} \]
Step 3: Determine the Mass of H₂O Produced
Now, we convert the mass of fat (1.2 kg) to grams: \[ 1.2 \text{ kg} = 1200 \text{ g} \]
Next, we calculate the number of moles of tristearin in 1200 g: \[ \text{Moles of tristearin} = \frac{1200 \text{ g}}{892.45 \text{ g/mol}} \approx 1.345 \text{ moles} \]
Then, using the balanced equation, for every mole of tristearin, 55 moles of H₂O are produced. Thus, the moles of H₂O produced: \[ \text{Moles of H}_2\text{O} = 1.345 \text{ moles of tristearin} \times 55 \text{ moles of H}_2\text{O/mole of tristearin} \approx 74.975 \text{ moles of H}_2\text{O} \]
Step 4: Convert Moles of H₂O to Mass
The molar mass of H₂O is: \[ \text{Molar mass of H}_2\text{O} = (2 \times 1.008) + (16.00) = 18.016 \text{ g/mol} \]
Now, calculate the mass of H₂O produced: \[ \text{Mass of H}_2\text{O} = 74.975 \text{ moles} \times 18.016 \text{ g/mol} \approx 1344.404 \text{ g} \]
Convert to kilograms: \[ \text{Mass of H}_2\text{O} = 1344.404 \text{ g} \approx 1.344 \text{ kg} \]
Final Answer
Thus, the mass of water produced is approximately 1.3 kg (to two significant figures).