-i- LPG stands for liquefied petroleum gas- Varieties of LPG are marketed including a mixture of propane -60-C3Hg-g- - 5O2-g- rarr- 3CO2-g- - 4H2O-g-2C4H10-g- - 13O2-g- rarr- 8CO2-g- - 10H2O-g-ii- Calculate the percentage of nitrogen and oxygen in ammonium nitrate -Relative molecular mass of ammonium nitrate is 80- H - 1- N - 14- O - 16-

(i) Volume of LPG = 10 L Volume of propane in 10 litres of LPG = 60100 #215;10 #160;= #160;6 #160;L Volume of butane in 10 litres of LPG = 40100 #215; ...

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Standard X

Chemistry

Mole Concept in Balanced Reaction

(i) LPG stand...

Question

(i) LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:
C₃H_g_(g) + 5O₂_(g) → 3CO₂_(g) + 4H₂O_(g)
2C₄H₁₀_(g) + 13O₂_(g) → 8CO₂_(g) + 10H₂O_(g)

(ii) Calculate the percentage of nitrogen and oxygen in ammonium nitrate [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16].

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Solution

(i) Volume of LPG = 10 L
Volume of propane in 10 litres of LPG = $\frac{60}{100} \times 10 = 6 L$
Volume of butane in 10 litres of LPG = $\frac{40}{100} \times 10 = 4 L$
During combustion, one mole of propane gives three moles of carbon dioxide.
Number of moles of carbon dioxide liberated from six litres of propane can be calculated from the ideal gas equation:
PV = nRT
Number of moles of propane in six litres of propane = $\frac{P \times 6}{R T}$
Number of moles of carbon dioxide liberated = $3 \times (\frac{P \times 6}{R T})$
Volume of carbon dioxide = $\frac{n R T}{P}$
Volume of carbon dioxide = $3 \times (\frac{P \times 6}{R T}) \times (\frac{R T}{P})$ = 18 L
Number of moles of carbon dioxide liberated from one mole of butane = 4
Volume of carbon dioxide liberated from butane is four times the volume of butane = 16 L
Total volume of carbon dioxide liberated = 18 + 16 = 34 L

(ii) Molecular mass of ammonium nitrate (NH₄)NO₃ = 80 g
Mass of nitrogen in ammonium nitrate = (14 + 14) g = 28 g
$Percentage of N in {NH}_{4} {NO}_{3} = \frac{Mass of N in compound}{Molecular mass of {NH}_{4} {NO}_{3}} \times 100 = \frac{28}{80} \times 100 = 35 %$

Mass of oxygen in ammonium nitrate = 3(16) g = 48 g
$Percentage of O in {NH}_{4} {NO}_{3} = \frac{Mass of O in compound}{Molecular mass of {NH}_{4} {NO}_{3}} \times 100 = \frac{48}{80} \times 100 = 60 %$

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Similar questions

LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:

(a) (i) LPG stands for liquefied petroleum gas. VArieties of LPG are marketed including a mixture fo propane (60%) and butane (40%) . If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions cab be represented as:
$C_{3} H_{8 (g)} + 50_{2 (g)} \to 3 C O_{2 (g)} + 4 H_{2} O_{(g)} 2 C_{4} H_{10 (g)} + 13 O_{2 (g)} \to 8 C O_{2 (g)} + 10 H_{2} O_{(g)}$
(ii)
Calculate the percentage of nitrogen and oxygen in ammonium nitrate. [Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16].
(b)4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.
(i) Write the equation for the reaction .
(ii) What is the mass of 4.5 moles of calcium carbonate ? (Relative molecular mas of calcium carbonate is 100).
(iii) What is the volume of carbon dioxide liberated at STP?
(iv) What mass of calcium chloride is formed?
(Relative molecular mass of calcium chloride is 111).
(v) How many moles of HCl are used in this reaction?

Q. If the relative molecular mass of ammonium nitrate is 80, the percentage of nitrogen and oxygen in ammonium nitrate is:

[N = 14, H = 1, O = 16]

Calculate the percentage of nitrogen and oxygen in ammonium nitrate.

[Relative molecular mass of ammonium nitrate is 80, H = 1, N= 14, 0= 16],

LPG ahs 60% propane and 40% butane ; 10 litres of this mixture is burnt. CAlculate the volume of carbonidioxide added to atmosphere.
$C_{3} H_{8} + 5 O_{2} \to 3 C O_{2} + 4 H_{2} O 2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O$

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