
In a Gobar gas plant, gobar gas is formed by bacterial fermentation of animal refuse. It mainly contains methane and its heat of combustion is -809 \[kJmo{l^{ - 1}}\] according to the following equation: $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$ ; $\Delta H$= -809 kJ
How much gobar gas would have to be produced per day for a small village of 50 families, if it is assumed that each family requires 20,000 kJ of energy per day? The methane content in gobar gas is 80% by mass.
Answer
487.2k+ views
Hint: First find out the total amount of methane that will be consumed by 50 families. Then find the amount of methane that would produce this much energy and amount of gobar gas associated with it.
Complete answer:
-According to the reaction: $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$ , the amount of energy evolved is 809 kJ.
-Now let us check the total amount of energy required by these families. It is given that one family requires 20,000kJ of energy and there are 50 families. So, the total amount of energy required is:
Total energy for 50 families = Consumption of energy by 1 family × number of families
= 20,000 × 50
= 10,00,000 kJ = ${10^6}$ kJ
-As we can see in the reaction that 809 kJ of energy is released when 1 mole of methane ($C{H_4}$) is burned. So we will use the unitary method to find out the amount of methane required to release ${10^6}$ kJ of energy.
809kJ of energy is evolved for : 1 mole of $C{H_4}$= 16 g $C{H_4}$
1 kJ of energy will be evolved for: $\dfrac{{16}}{{809}}$ g of $C{H_4}$
${10^6}$ kJ of energy will be evolved by: $\dfrac{{16}}{{809}} \times {10^6}$ g of $C{H_4}$
= 19,770 g = 19.77 kg of methane
-The question says that the methane content in gobar gas is 80% by mass.
So, let the amount of gobar gas that contains 19.77 kg of methane be ‘x’ kg. So, we can say that 80% of ‘x’ is 19.77 kg.
$\dfrac{{80}}{{100}} \times x = 19.77$
$x = 19.77 \times \dfrac{{100}}{{80}}$
X = 24.7125 kg gobar gas
So, 24.7125 kg of gobar gas will be required by 50 families.
Note: Gobar gas is mainly composed of methane (\[C{H_4}\]), carbon dioxide ($C{O_2}$), small amounts of hydrogen sulphide (${H_2}S$) and trace amounts of nitrogen (${N_2}$), hydrogen (${H_2}$) and carbon monoxide ($CO$). This gas is highly flammable due to the presence of methane and produces blue flame on burning. Thus, it is used as a source of energy. 25 times more heat can be captured by one pound of methane than by one pound of carbon dioxide.
Complete answer:
-According to the reaction: $C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O$ , the amount of energy evolved is 809 kJ.
-Now let us check the total amount of energy required by these families. It is given that one family requires 20,000kJ of energy and there are 50 families. So, the total amount of energy required is:
Total energy for 50 families = Consumption of energy by 1 family × number of families
= 20,000 × 50
= 10,00,000 kJ = ${10^6}$ kJ
-As we can see in the reaction that 809 kJ of energy is released when 1 mole of methane ($C{H_4}$) is burned. So we will use the unitary method to find out the amount of methane required to release ${10^6}$ kJ of energy.
809kJ of energy is evolved for : 1 mole of $C{H_4}$= 16 g $C{H_4}$
1 kJ of energy will be evolved for: $\dfrac{{16}}{{809}}$ g of $C{H_4}$
${10^6}$ kJ of energy will be evolved by: $\dfrac{{16}}{{809}} \times {10^6}$ g of $C{H_4}$
= 19,770 g = 19.77 kg of methane
-The question says that the methane content in gobar gas is 80% by mass.
So, let the amount of gobar gas that contains 19.77 kg of methane be ‘x’ kg. So, we can say that 80% of ‘x’ is 19.77 kg.
$\dfrac{{80}}{{100}} \times x = 19.77$
$x = 19.77 \times \dfrac{{100}}{{80}}$
X = 24.7125 kg gobar gas
So, 24.7125 kg of gobar gas will be required by 50 families.
Note: Gobar gas is mainly composed of methane (\[C{H_4}\]), carbon dioxide ($C{O_2}$), small amounts of hydrogen sulphide (${H_2}S$) and trace amounts of nitrogen (${N_2}$), hydrogen (${H_2}$) and carbon monoxide ($CO$). This gas is highly flammable due to the presence of methane and produces blue flame on burning. Thus, it is used as a source of energy. 25 times more heat can be captured by one pound of methane than by one pound of carbon dioxide.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

The correct geometry and hybridization for XeF4 are class 11 chemistry CBSE

Water softening by Clarks process uses ACalcium bicarbonate class 11 chemistry CBSE

With reference to graphite and diamond which of the class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
