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Mandarinka [93]
3 years ago
6

The average propane cylinder for a residential grill holds approximately 18 kg of propane. how much energy (in kj) is released b

y the combustion of 20.00 kilograms of propane in sufficient oxygen?
Chemistry
1 answer:
Angelina_Jolie [31]3 years ago
6 0
Let's begin with the basic values  that will be used in the solution.

The formula of propane is C3H8. It is an alkane, a hydrocarbon with the general formula of CnH2n+2. Notice that hydrocarbons have only Carbon and Hydrogen atoms. Its molar mass (M) is 44 g.

Molar Mass Calculation is done as like that
C=12 g/mol, H=1 g/mol. 1 mole propane has 3 moles Carbon atoms and 8 mole Hydrogen atoms. M(
C3H8)= 3*12+ 8*1= 44 g

Combustion reaction of hydrocarbons gives carbon dioxide and water by releasing energy. That energy is called as enthalpy of combustion (
ΔHc°).  

ΔHc° of propane equals -2202.0 kj/mol. Burning of 1 mole C3H8 releases 2202 kj energy. Minus sign only indicates that the energy is given out ( an exothermic reaction ).

Let's write the combustion reaction.
C3H8 + O2 ---> CO2 + H20 (unbalanced) 
ΔHc° = -2202 kj/mol

Now, we calculate mole of 20 kg propane. Convert kilogram into gram since we use molar mass is defined in grams.
mole=mass/molar mass ; n=m/M ; n= 20000 g /44 (g/mol)=454 mole

1 mole propane releases 2202 kj energy.
454 mole propane release 2202 kj *454= 1000909 kj

The answer is 1000909 kj.



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Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

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Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

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