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Mole Concept & Avogadro Number

The fat, glyc...

Question

The fat, glyceryl trioleate, is metabolized via the following reaction. Given the enthaplies of formation, calculate the energy (kJ) liberated when 1.00 g of this fat reacts.
(Atomic weights : $C = 12.01, H = 1.008, O = 16.00$ ).
$C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) ⟶ 57 C O_{2} (g) + 52 H_{2} O (I)$
$Δ H C_{57} H_{107} O_{6} = - 70870 k J / m o l e$
$Δ H H_{2} O (I) = - 285.8 k J / m o l e$
$Δ H C O_{2} (g) = - 393.5 k J / m o l e$

40.4

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33.4

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37.8

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42.6

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Solution

The correct option is C 37.8
$Δ H_{r}$ = 57X $Δ H C O_{2} (g)$ + 52X $Δ H H_{2} O (I)$ - $Δ H C_{57} H_{107} O_{6}$
$Δ H_{r}$ = -14861.6 -22429.5 +70870 = 33578.9 kJ/mol
Molar mass of $C_{57} H_{107} O_{6}$ = 888.402 g/mol
when 888.402 g of $C_{57} H_{107} O_{6}$ is metabolized $Δ H_{r}$ = 33578.9 kJ
Therefore, when 1g of $C_{57} H_{107} O_{6}$ is metabolized $Δ H_{r}$ = 33578.9/888.402 = 37.8 kJ

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Similar questions

The fat $C_{57} H_{104} O_{6}$ (s) is metabolized via the following reaction.
calculate the energy (KJ) liberated when 1g of this fat reacts.

$C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) \to 57 C O_{2} (g) + 52 H_{2} O (l)$

Given the enthalpies of formation:

$Δ H_{f}^{0} (C_{57} H_{104} O_{6}, s) = - 70870$ kJ/mol
$Δ H_{f}^{0} (H_{2} O, l) = - 285.8$ kJ/mol
$Δ H_{f}^{0} (C O_{2}, g) = - 393.5$ kJ/mol

C_{57} H_{104} O_{6} (s)

C_{57} H_{104} O_{6} (s) + 80 O_{2} (g) \to 57 C O_{2} (g) + 52 H_{2} O (I)

Δ H_{f}^{0} (C_{57} H_{104} O_{6}, s) = - 70870 k J / m o l Δ H_{f}^{0} (H_{2} O, l) = - 285.8 k J / m o l Δ H_{f}^{0} (C O_{2}, g) = - 393.5 k J / m o l

3.00 g

N H_{3 (g)}

N = 14.00, H = 1.008

4 N H_{3} (g) + 5 O_{2} (g) ⟶ 4 N O (g) + 6 H_{2} O (g)

Δ H N H_{3} (g) = - 46.1 k J / m o l e

Δ H N O (g) = + 90.2 k J / m o l e

Δ H H_{2} O (g) = - 241.8 k J / m o l e

C_{3} H_{4}

C_{3} H_{4} (g) + 4 O_{2} (g) \to 3 C O_{2} (g) + 2 H_{2} O (l)

Δ H^{\circ} = - 1939.1 k J

C O_{2 (g)}

H_{2} O_{J}

Δ H f

C O_{2 (g)}

H_{2} O_{J}

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