To calculate the mass of water (H₂O) produced from the metabolism of 1.2 kg of tristearin (C₅₇H₁₁₀O₆), we will follow these steps:
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Determine the molar mass of tristearin (C₅₇H₁₁₀O₆).
- Carbon (C): 12.01 g/mol × 57 = 684.57 g/mol
- Hydrogen (H): 1.008 g/mol × 110 = 110.88 g/mol
- Oxygen (O): 16.00 g/mol × 6 = 96.00 g/mol
Molar mass of tristearin = 684.57 + 110.88 + 96.00 = 891.45 g/mol
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Calculate the number of moles of tristearin in 1.2 kg (1200 g).
\[ \text{Moles of tristearin} = \frac{1200 , \text{g}}{891.45 , \text{g/mol}} \approx 1.344 , \text{mol} \]
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Write the balanced equation for the metabolism of tristearin:
The balanced metabolic equation is:
\[ C_{57}H_{110}O_6 + O_2 \rightarrow CO_2 + H_2O \]
The complete balanced equation will be:
\[ C_{57}H_{110}O_6 + 86O_2 \rightarrow 57CO_2 + 55H_2O \]
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Determine the moles of H₂O produced:
From the equation, 1 mole of C₅₇H₁₁₀O₆ produces 55 moles of H₂O. Thus:
\[ \text{Moles of H₂O produced} = 1.344 , \text{mol} \times 55 = 73.92 , \text{mol} \]
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Calculate the mass of H₂O produced:
The molar mass of H₂O is:
- Hydrogen (H): 1.008 g/mol × 2 = 2.016 g/mol
- Oxygen (O): 16.00 g/mol × 1 = 16.00 g/mol
Molar mass of H₂O = 2.016 + 16.00 = 18.016 g/mol
Now we can find the total mass of H₂O produced:
\[ \text{Mass of H₂O} = 73.92 , \text{mol} \times 18.016 , \text{g/mol} \approx 1333.67 , \text{g} \]
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Convert to kg and round to two significant figures:
\[ 1333.67 , \text{g} = 1.334 , \text{kg} \]
Rounded to two significant figures, the mass of H₂O produced is:
\[ \boxed{1.3} , \text{kg} \]