Chemistry, 15.10.2019 17:30 heyItsLuna234

At a festival, spherical balloons with a radius of 280. cm are to be inflated with hot air and released. the air at the festival will have a temperature of 25° c and must be heated to 100°c to make the balloons float. 3.00 kg of propane (c3h8) fuel are available to burn to heat the air. calculate the number of balloons that can be inflated with hot air.

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Answer from: Deagon83951

The question is incomplete, here is the complete question:

At a festival, spherical balloons with a radius of 280 cm are to be inflated with hot air and released. The air at the festival will have a temperature of 25° C and must be heated to 100°C to make the balloons float. 3.00 kg of propane (C₃H₈) fuel are available to burn to heat the air. Calculate the number of balloons that can be inflated with hot air.

Density of air at 100°C = 0.946kg/m^3

Specific heat capacity of air = 1.009 J/g°C

The number of balloons inflated with hot air are 7.

Explanation:

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

The chemical equation for the combustion of propane follows:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(H_2O(g))})]-[(1\times \Delta H_f_{(C_3H_8(g))})+(5\times \Delta H_f_{(O_2(g))})]$

We are given:

$\Delta H_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.51kJ/mol\\\Delta H_f_{(C_3H_8(g))}=-103.8kJ/mol\\\Delta H_f_{(O_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-393.51))+(4\times (-241.8))]-[(1\times (-103.8))+(5\times (0))]\\\\\Delta H_{rxn}=-2043.93kJ/mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of propane = 3.00 kg = 3000 g (Conversion factor: 1 kg = 1000 g)

Molar mass of propane = 44 g/mol

Putting values in above equation, we get:

$\text{Moles of propane}=\frac{3000g}{44g/mol}=68.18mol$

To calculate the heat released, we use the equation:

$\Delta H_{rxn}=\frac{q}{n}$

where,

q = amount of heat released = ? kJ

n = number of moles = 68.18 moles

$\Delta H_{rxn}$ = enthalpy change of the reaction = -2043.93 kJ/mol

Putting values in above equation, we get:

$-2043.93kJ/mol=\frac{q}{68.18mol}\\\\q=(-2043.93kJ/mol\times 68.18mol)=-1.39\times 10^5kJ$

To calculate the heat released by the reaction, we use the equation:

$q=mc\Delta T$

where,

q = heat released = $1.39\times 10^5kJ=1.39\times 10^8J$

m = mass of air = ?

c = heat capacity of air = 1.009 J/g°C

$\Delta T$ = change in temperature = T_2-T_1=(100-25)^oC=75^oC

Putting values in above equation, we get:

$1.39\times 10^8J=m\times 1.009J/g^oC\times 75^oC\\\\m=\frac{1.39\times 10^8}{1.009\times 75}=1.84\times 10^6g=1.84\times 10^3kg$

Conversion factor: 1 kg = 1000 g

To calculate the volume of air, we use the equation:

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

Density of air = 0.946kg/m^3

Mass of air = $1.84\times 10^3kg$

Putting values in above equation, we get:

$0.946kg/m^3=\frac{1.84\times 10^3kg}{\text{Volume of air}}\\\\\text{Volume of air}=\frac{1.84\times 10^3kg}{0.946kg/m^3}=1945.03m^3$

To calculate the volume of sphere, we use the equation:

$V=4\pi r^3$

where,

r = radius of the sphere = 280 cm = 2.8 m (Conversion factor: 1 m = 100 cm)

Putting values in above equation, we get:

$V=4\times 3.14\times (2.8)^3\\\\V=275.7m^3$

To calculate the number of balloons inflated, we divide the volume of air by the volume of 1 balloon:

$\text{Number of balloons inflated}=\frac{V_{air}}{V_{balloon}}\\\\\text{Number of balloons inflated}=\frac{1945.03m^3}{275.7m^3}=7$

Hence, the number of balloons inflated with hot air are 7.

Answer from: Quest

pie chart because it is in percents ( % )

Answer from: Quest

The answer is !

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