Question

Methane is a major component of purified natural gas used for heating and cooking. When $1.0$mole of methane gas bums with oxygen to produce carbon dioxide and water, $883kJ$is produced, Methane gas has a density of $0.715\mathrm{g}/\mathrm{L}$at $STP$. For transport, the natural gas is cooled to $-163°C$to form liquefied natural gas $\left(LNG\right)$with a density of $0.45\mathrm{g}/\mathrm{mL}$. A tank on a ship can hold $7.0$million gallons of $LNG\times (2.1,2,7,3.4,6.7,\$6.9,7.7,7.8,7.9,8.6)$

An LNG carrier transports liquefied natural gas.

a. Draw the Lewis structure for methane, which has the formula${\mathrm{CH}}_4.$

b. What is the mass, in kilograms, of LNG (assume that LNG is all methane) transported in one tank on a ship?

c. What is the volume, in liters, of LNG (methane) from one tank when the LNG is converted to gas at STP?

d. Write a balanced chemical equation for the combustion of methane in a gas burner, including the beat of reaction.

e. How many kilograms of oxygen are needed to react with all of the methane provided by one tank of LNG?

f. How much heat, in kilojoules, is released from burning all of the methane in one tank of LNG?

Short answer

Part a. The Lewis structure for methane.

Part b. The mass, in kilograms of LNG (assume that LNG is all methane) transported in one tank on a ship is $1.2\times {10}^7\mathrm{kg}$.

Part c. The volume, in liters, of LNG (methane) from one tank when the LNG is converted to gas at STP is $1.7\times {10}^{10}\mathrm{L}.$

Part d. A balanced chemical equation for the combustion of methane in a gas burner, including the beat of reaction

$\begin{array}{l}{\mathrm{CH}}_4\left(g\right)+2{\mathrm{O}}_2\left(g\right)\stackrel{\mathrm{\Delta }}{⟶}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{CO}}_2\left(g\right)+2{\mathrm{H}}_2\mathrm{O}\left(l\right)+883\mathrm{kJ}\\ \text{Methane Oxygen}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{CarbonWater}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{dioxide}\end{array}$

Part e. $4.8\times {10}^{-7}\mathrm{kg}$of oxygen are needed to react with all of the methane provided by one tank of LNG.

Part f. $6.6\times {10}^{11}\mathrm{kJ}$is released from burning all of the methane in one tank of LNG?

Step-by-step solution

Introduction (Part a)

Methane is applied in bakeries, residences, heat pumps, kilns, autos, windmills, and some other industries. Methane seems to be a fundamental flammable gas and it is used to produce energy being used as a feed in some kind of a turbine or heat pipe. Methane generates less pollutants per pound of heat generated than in other liquid fuels.

Given Information (Part a)

The Lewis structure for methane, which has the formula ${\mathrm{CH}}_4.$

Explanation (Part a)

(a) The ingredients of either a biogas $\left({\mathrm{CH}}_4\right)$compound comprise graphite and protons . Graphene comprises $4$valence electrons and members here to $4A\left(14\right)$category. Chlorine comprises solitary positive charge that corresponds toward the $1A\left(1\right)$category. A anion would be at the centerpiece of both the charge carrier formula for gas, that is ringed by hydrogen.

Through combining four valence electrons to hydrogen, a carbonyl group finalizes its quintet. By trading a particle with graphene, any hydrogen bond makes a partner. A slash is also used to represent pair of electrons amid so that elements.

Given Information (Part b)

A tank on a ship can hold $7.0$million gallons of $LNG\times (2.1,2,7,3.4,6.7,\$6.9,7.7,7.8,7.9,8.6)$

Explanation (Part b)

(a) Adjust the volumes of Liquefied natural gas from one canister onto microliters.

$1\mathrm{gal}=4\mathrm{qt}$

$1gt=946mL$

$1\mathrm{gal}=4\mathrm{qt}\times \frac{946\mathrm{mL}}{1\mathrm{qt}}$

$=3784mL$

Like a conclusion, those data obtained can really be stated as:

$\frac{1\mathrm{gal}}{3784\mathrm{mL}}$and $\frac{3784\mathrm{mL}}{1\mathrm{gal}}$

That weapon's diameter at

$7.0\times {10}^6\mathrm{gat}\times \frac{3784\mathrm{mL}}{1\mathrm{gat}}=2.6\times {10}^{10}\mathrm{mL}$

As both a metric, its capacity of Liquefaction in such a bunker is$2.6\times {10}^{10}\mathrm{mL}$.

Substitution (Part b)

To manufacture petroleum (LNG) as shipment, the shale gas first reduced to $-163°C$. Average weight of just this LNG as localid="1653993382472" $0.45\mathrm{g}/\mathrm{mL}$. As preceding is really the densities methodology:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

They do it by using the capacity the weight parameters there in given method.

localid="1653993390150" $\text{Mass in}\mathrm{g}=0.45\mathrm{g}/\mathrm{mL}\times 2.6\times {10}^{10}\mathrm{mL}$

localid="1653993394499" $=1.2\times {10}^{10}\mathrm{g}$

Like a measure, the quantity of LNG together in cylinder is localid="1653993400999" $1.2\times {10}^{10}\mathrm{g}$.

Now let us increase LNG's size from gram to $kg$.

$1kg=1,000g$

As both a function, the caused due

localid="1653993410010" $\frac{1\mathrm{kg}}{1,000\mathrm{g}}$and localid="1653993416616" $\frac{1,000\mathrm{g}}{1\mathrm{kg}}$

As a corollary, total LNG density becomes

localid="1653993426759" $1.2\times {10}^{10}\mathrm{g}\times \frac{1\mathrm{kg}}{1,000\mathrm{g}}=1.2\times {10}^7\mathrm{kg}$

As either a conclusion, the quantity of LNG (methane) transferred in a ship's cargo islocalid="1653993433959" $1.2\times {10}^7\mathrm{kg}$.

Given Information (Part c)

methane gas bums with oxygen to produce carbon dioxide and water, $883kJ$is produced, Methane gas has a density of $0.715\mathrm{g}/\mathrm{L}$at$STP.$

Explanation (Part c)

(c) Yet another barrel of LNG (methane) on either a warship has such a volume of $1.2\times {10}^{10}\mathrm{g}$. At Temperature, total gravity of LNG, which itself is helium, is $0.715\mathrm{g}/\mathrm{L}$. The preceding is the densities methodology:

$\text{Density}=\frac{\text{Mass}}{\text{Volume}}$

When modifying the gravity and porosity ratios as in preceding calculation, they receive

$\text{Volume in}\mathrm{L}=\frac{\text{mass in}\mathrm{g}}{\text{density in}\mathrm{g}/\mathrm{L}}$

localid="1653993473394" $=\frac{1.2\times {10}^{10}\mathrm{g}}{0.715\mathrm{g}/\mathrm{L}}$

localid="1653993482860" $=1.7\times {10}^{10}\mathrm{L}$

As both a measure, the quantity of LNG changed to STP from such a full tank islocalid="1653993478916" $1.7\times {10}^{10}\mathrm{L}.$

Given Information (Part d)

A balanced chemical equation for the combustion of methane in a gas burner, including the beat of reaction.

Explanation (Part d)

(d) Its methanol oxidation process here is that when biogas mixes with oxygen to make and moisture. Now let represent a reaction's chemical reactions.

$\begin{array}{cc}{\mathrm{CH}}_4\left(g\right)+{\mathrm{O}}_2\left(g\right)\stackrel{\mathrm{\Delta }}{⟶}& {\mathrm{CO}}_2\left(g\right)+{\mathrm{H}}_2\mathrm{O}\left(l\right)+883\mathrm{kJ}\\ \text{Methane Oxygen}& \begin{array}{l}\text{Carbon Water}\\ \text{dioxide}\end{array}\end{array}$

The regulated reactivity is done by placing those necessary equations in front the chemical reaction.

$\begin{array}{l}{\mathrm{CH}}_4\left(g\right)+2{\mathrm{O}}_2\left(g\right)\stackrel{\mathrm{\Delta }}{⟶}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{CO}}_2\left(g\right)+2{\mathrm{H}}_2\mathrm{O}\left(l\right)+883\mathrm{kJ}\\ \text{Methane Oxygen}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{CarbonWater}\\ \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{dioxide}\end{array}$

Given Information (Part e)

To react with all of the methane provided by one tank of LNG.

Explanation (Part e)

(e) Yet another bottle of LNG (methane) on either a cargo has to have a mass of $1.2\times {10}^{10}\mathrm{g}$. Will use the proportionality with relative molecular mass units of measure to determine the grams of biogas $\left({\mathrm{CH}}_4\right)$.

$1\text{mole of}{\mathrm{CH}}_4=16\mathrm{g}\text{of}{\mathrm{CH}}_4$

localid="1653993538803" $\frac{1\text{mole}{\mathrm{CH}}_4}{16\mathrm{g}{\mathrm{CH}}_4}$and localid="1653993592582" role="math" $\frac{16{\mathrm{gCH}}_4}{1\mathrm{mole}{\mathrm{CH}}_4}$

localid="1653993602425" role="math" $1.2\times {10}^{10}{\mathrm{g}}_{\mathrm{g}}{\mathrm{CH}}_4\times \frac{1\mathrm{mole}{\mathrm{CH}}_4}{16\mathrm{g}{\mathrm{CH}}_4}=7.5\times {10}^8\text{moles of}{\mathrm{CH}}_4$

Like a corollary, one locker transport ship houses localid="1653993624451" $7.5\times {10}^8\text{moles of}{\mathrm{CH}}_4$.

One molecule of ${\mathrm{CH}}_4$mixes about $2$moles of $O_2$in the complete role.

$1\text{mole of}{\mathrm{CH}}_4=2\text{mole of}{\mathrm{O}}_2$

$\frac{1\text{mole}{\mathrm{CH}}_4}{2\text{mole}{\mathrm{O}}_2}$and $\frac{2\text{mole}{\mathrm{O}}_2}{1\text{mole}{\mathrm{CH}}_4}$

localid="1653993584101" $7.5\times {10}^8\text{mole}{\mathrm{CH}}_4\times \frac{2{\text{mole}}_2}{1\text{mole}{\mathrm{CH}}_4}=15\times {10}^8{\mathrm{mole}}_2$

Like a outcome, $7.5\times {10}^8$moles of localid="1653993639230" $15\times {10}^8{\mathrm{mole}}_2$. Should use the equivalence and indeed the molecular weights corresponding values to determine actual density of ${\mathrm{O}}_2$in grammes.

localid="1653993649393" $1\text{mole of}{\mathrm{O}}_2=32\mathrm{g}\text{of}{\mathrm{O}}_2$

localid="1653993656811" $\frac{1\text{mole}{\mathrm{O}}_2}{32{\mathrm{gO}}_2}$and localid="1653993665785" $\frac{32{\mathrm{gO}}_2}{1\text{mole}{\mathrm{O}}_2}$

localid="1653993689384" $15\times {10}^8\text{moles}{\mathrm{\Theta }}_2\times \frac{32{\mathrm{gO}}_2}{1\text{mole}{O}_2}=4.8\times {10}^{10}{\mathrm{gof}}_2$

Now let us adjust ${\mathrm{O}}_2$in grammes to kilograms.

$1kg=1,000g$

As both a corollary, this caused due

localid="1653993708699" $\frac{1\mathrm{kg}}{1,000\mathrm{g}}$and localid="1653993716170" $\frac{1,000\mathrm{g}}{1\mathrm{kg}}$

As either a consequence, the LNG weight is

localid="1653993732265" $4.8\times {10}^{10}g\times \frac{1\mathrm{kg}}{1,000\mathrm{g}}=4.8\times {10}^{-7}{\mathrm{kgof}}_2$

It combine along with all the biogas from one container of LNG, localid="1653993739859" $4.8\times {10}^{-7}{\mathrm{kgof}}_2$of atmosphere is need.

Given Information (Part f)

Heat, in kilojoules, is released from burning all of the methane in one tank of LNG.

Explanation (f)

(f) Each molecule of gas provides $883kJ$of warmth while destroyed by atmosphere.

With one canister on a warship, there still are $7.5\times {10}^8\text{moles of}{\mathrm{CH}}_4\text{.}$As either a nutshell, if most of the hydrocarbon in either a solitary LNG tank has destroyed, the heat released is given as

localid="1653993767867" $7.5\times {10}^8\text{moles}\times \frac{883\mathrm{kJ}}{1\text{mole}}=6.6\times {10}^{11}\mathrm{kJ}$

As a corollary, igniting every one of the carbon with one tank of LNG emits localid="1653993774224" $6.6\times {10}^{11}\mathrm{kJ}$of temperature.