To calculate the mass of water (H₂O) produced from the metabolism of 1.2 kg of tristearin (C₅₇H₁₁₀O₆), we first need to go through the balanced chemical reaction and then convert the mass of fat into moles. Here are the steps involved:
Step 1: Write the Balanced Equation
The combustion of tristearin (C₅₇H₁₁₀O₆) can be represented as: \[ \text{C}{57}\text{H}{110}\text{O}{6} + \text{O}{2} \rightarrow \text{CO}{2} + \text{H}{2}\text{O} \]
Balancing the equation, we have: \[ \text{C}{57}\text{H}{110}\text{O}{6} + 83\text{O}{2} \rightarrow 57\text{CO}{2} + 55\text{H}{2}\text{O} \]
Step 2: Calculate Molar Mass of Tristearin
The molar mass of tristearin (C₅₇H₁₁₀O₆) can be calculated as follows:
- Carbon (C): \( 57 \times 12.01 , \text{g/mol} \)
- Hydrogen (H): \( 110 \times 1.008 , \text{g/mol} \)
- Oxygen (O): \( 6 \times 16.00 , \text{g/mol} \)
Calculating each component: \[ \text{C:} , 57 \times 12.01 = 685.57 , \text{g/mol} \] \[ \text{H:} , 110 \times 1.008 = 110.88 , \text{g/mol} \] \[ \text{O:} , 6 \times 16.00 = 96.00 , \text{g/mol} \]
Adding these together gives: \[ \text{Molar mass of C}{57}\text{H}{110}\text{O}_{6} = 685.57 + 110.88 + 96.00 = 892.45 , \text{g/mol} \]
Step 3: Convert Mass of Tristearin to Moles
Next, we convert the given mass of fat (1.2 kg) to grams: \[ 1.2 , \text{kg} = 1200 , \text{g} \]
Now, we can convert grams to moles: \[ \text{Moles of tristearin} = \frac{1200 , \text{g}}{892.45 , \text{g/mol}} \approx 1.344 , \text{mol} \]
Step 4: Use Stoichiometry to Find Moles of Water Produced
From the balanced equation, we see that 1 mole of tristearin produces 55 moles of H₂O. Thus: \[ \text{Moles of H}2\text{O} = 1.344 , \text{mol C}{57}\text{H}{110}\text{O}{6} \times \frac{55 , \text{mol H}2\text{O}}{1 , \text{mol C}{57}\text{H}{110}\text{O}{6}} \approx 73.92 , \text{mol H}_2\text{O} \]
Step 5: Convert Moles of Water to Mass
The molar mass of water (H₂O) is: \[ 2 \times 1.008 , \text{g/mol} + 16.00 , \text{g/mol} = 18.016 , \text{g/mol} \]
Now, convert moles of water to grams: \[ \text{Mass of H}_2\text{O} = 73.92 , \text{mol} \times 18.016 , \text{g/mol} \approx 1333.23 , \text{g} \]
Step 6: Convert to Kilograms and Round to Two Significant Figures
To express this in kilograms: \[ 1333.23 , \text{g} = 1.33323 , \text{kg} \]
Rounded to two significant figures gives: \[ \boxed{1.3} , \text{kg} \]
So, the mass of H₂O produced by the metabolism of 1.2 kg of fat is approximately 1.3 kg.