using question 21 and subsequent answers I got lost from e to h 21) When one mol

Question

using question 21 and subsequent answers I got lost from e to h

21) When one mole of methane gas is combusted with oxygen, 883 kJ ls released. Methane gas has a de of 0.715 g/l at STP. when transported the gas is cooled to-163°C to form liquefied natural gas (LNG) which has a density of 0.45 g/mL. A tank on a large transport ship contain 7.0 million galions of LNG (8pts) a) Draw the Lewis sructure for methane. CH b) What is the mass (kg) of LNG that is contained in one tank on a transport ship? c) What is the volume (U that this gas would occupy at STP conditions? I D d) Write a balanced reaction for the combustion of methane. 833

Explanation / Answer

Ans. #b. Volume of LNG = 7.0 x 106 gallons = 2.6498 x 107 L

Now,

            Mass of LNG = Vol. x Density of LNG = 2.6498 x 107 L x (0.45 kg/ L)

                                    = 1.1924 x 107 kg

#c. Vol. of CH4 at STP = Mass in grams / Density at STP

                                    = 1.1924 x 1010 g / (0.715 g / L)

                                    = 1.6677 x1010 L

#e. Moles of CH4 in tank = Mass / MW

                                    = 1.1924 x 1010 g / 16.0 g mol-1 = 7.4525 x 108 mol

Following stoichiometry, 1 mol CH4 requires 2 mol O2 for complete combustion.

So,

            Required moles of O2 = 2 x Moles of CH4 to be combusted

                                                = 2 x 7.4525 x 108 mol = 1.4905 x 109 mol

# Now, required mass of O2 = Required moles of O2 x MW

                                                = 1.4905 x 109 mol x (32.0 g/ mol)

                                                = 4.7696 x 1010 g

                                                = 4.7696 x 107 kg

#f. Amount of total heat released = Moles of CH4 x Molar enthalpy of CH4 combustion                                        = 7.4525 x 108 mol x (-833.0 kJ/ mol)

                                                = 6.2080 x 1011 kJ

#g. Heat change of a substance is given by-

q = m s dT                            - equation 1

Where,

q = heat change

m = mass

s1 = specific heat

dT = Final temperature – Initial temperature = (T2 – T1)0C

Mass of water sample= 3.0 x 109 kg

Putting the values in equation 1-

            6.2080 x 1011 kJ = 3.0 x 109 kg x (4.184 kJ kg-1 0C-1) x (T2 – 1.00C)

Or, T2 – 1.00C = 6.2080 x 1011 kJ / (1.2552 x 1010 kJ 0C-1) = 49.460C

Hence, T2 = 49.460C + 1.000C = 50.460C

Therefore, the final temperature, T2 = 50.460C

#h. Volume = Mass / density

                        = 3.0 x 109 kg / (1.0 g/ mL)

                        = 3.0 x 109 kg / (1.0 kg/ L)

                        = 3.0 x 109 L

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