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Chapter 3: Problem 101

The fat stored in a camel's hump is a source of both energy and water. Calculate the mass of \(\mathrm{H}_{2} \mathrm{O}\) produced by the metabolism of \(1.0 \mathrm{~kg}\) of fat, assuming the fat consists entirely of tristearin \(\left(\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}\right)\), a typical animal fat, and assuming that during metabolism, tristearin reacts with \(\mathrm{O}_{2}\) to form only \(\mathrm{CO}_{2}\) and $\mathrm{H}_{2} \mathrm{O}$.

Short Answer

Expert verified
The mass of water produced by the metabolism of 1.0 kg of tristearin fat is approximately 1.11 kg.

Step by step solution

01

Determine the molar mass of tristearin

Knowing that tristearin has a molecular formula of C57H110O6, we can calculate its molar mass by using the atomic masses of carbon, hydrogen, and oxygen: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of O = 16.00 g/mol The molar mass of tristearin is thus: \(57(12.01) + 110(1.01) + 6(16.00) = 891.27 g/mol\)
02

Convert 1 kg of fat to moles

Now that we know the molar mass, we can convert the given mass of 1.0 kg tristearin to moles: (1000 g fat) x (1 mol tristearin/891.27 g tristearin) = \(1.122 \text{ moles of tristearin}\)
03

Write and balance the chemical equation for the metabolism of tristearin

The balanced chemical equation for the metabolism of tristearin is: \(C_{57}H_{110}O_{6} + 82O_{2} \rightarrow 57CO_{2} + 55H_{2}O\)
04

Calculate moles of water produced from 1 kg of tristearin

Using stoichiometry, we can determine the moles of water produced from the 1.122 moles of tristearin: \(1.122 \text{ moles of tristearin} \times \dfrac{55 \text{ moles of water}}{1 \text{ mole of tristearin}} = 61.71 \text{ moles of water}\)
05

Convert moles of water to mass

Finally, we convert the moles of water produced to mass using the molar mass of water: Molar mass of water = (2 x 1.01 g/mol) + (1 x 16 g/mol) = 18.02 g/mol \(61.71 \text{ moles of water} \times \dfrac{18.02 \text{g of water}}{1 \text{ mole of water}} = 1112.24 \text{g of water}\) Therefore, the mass of water produced by metabolism of 1.0 kg of tristearin fat is approximately 1112.24 g or 1.11 kg.

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Most popular questions from this chapter

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) $\mathrm{C}_{7} \mathrm{H}_{16}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ (b) $\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)$ (c) $\mathrm{Zn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(I)$ (d) $\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Ag}(s)+\mathrm{O}_{2}(g)$

Balance the following equations: (a) $\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)$ (b) $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)$ (d) $\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)$

The allowable concentration level of vinyl chloride, $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl},\( in the atmosphere in a chemical plant is \)2.0 \times 10^{-6} \mathrm{~g} / \mathrm{L}$. How many moles of vinyl chloride in each liter does this represent? How many molecules per liter?

A piece of aluminum foil \(1.00 \mathrm{~cm}^{2}\) and \(0.550-\mathrm{mm}\) thick is allowed to react with bromine to form aluminum bromide. (a) How many moles of aluminum were used? (The density of aluminum is $2.699 \mathrm{~g} / \mathrm{cm}^{3} .$ ) (b) How many grams of aluminum bromide form, assuming the aluminum reacts completely?
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