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.
I would say compression or the absence of heat because heating a liquid makes it a gas<span />
Limiting reactant in this experiment would be Magnesium since it will run out first
Answer:
3,200 joules
Explanation:
q = mcΔT = (250.0 g)(0.128 j/g°C)(100°C -25°C) = 3,200 joules
