How much energy in kilocalories would combustion of 0.740 ounces of methane release?

Question

I need to express it in 3 significant figures

Answer 1

Looking up the ∆H_combustion for methane I find it to be -800 kJ/mole.

Converting 0.740 ounces to grams and then to moles of methane, we have...

0.740 oz x 1 lb / 16 oz x 454 g / lb = 20.998 g methane

20.998 g methane x 1 mol methane / 16 g = 1.312 mols methane

-800 kJ/mol x 1.312 mols = -1050 kJ

-1050 kJ x 1 kcal / 4.184 kJ = -251 kcal (3 sig. figs.)

Answer 2

My answer may differ depending on what bond dissociation energy table is used for your course.

You need to 1) calculate the energy difference between the products and the reagents for the balanced equation for the combustion of methane. 2) Then multiply that energy by the calculated moles of methane of 0.740 ounces.

1)

Equation: CH₄ + 2 O₂ = CO₂ + 2H₂O

Bonds broken: 4 C-H bonds (337.2 kJ), 2*2 O=O bonds (498.34 kJ)

Bonds formed: 2 C-O bonds (749 kJ) , 2*H-O bonds (428 kJ)

Energy of Bonds broken = 4*337.2 + 2*2*498.34

Energy of bonds formed = 2*749 + 2*2*428

Energy Bonds broken - Bonds formed = -864.52 kJ/mole

Conversion: kJ -> kCal (4.184 kJ in 1 kCal) -> -206.63 kCal

2) Calculate 0.74 oz to moles of methane

using 16 oz in 1lb; 453.59 g in 1 lb and MW 16g in 1 mole

I got 1.311 moles of methane

Answer: 1.311 * -206.63 = -270.88 kCal -> -271 kCal