Calculate Heat in Combustion of C4H10(g)

[needhelp83]

How much heat, in kilojoules, is evolved in the complete combustion of 1.325 g C₄H₁₀(g) at 23.6 degrees Celsius and 738 mmHg? The complete combustion of butane is represented by this equation

C₄H₁₀(g) + $$\frac{13}{2}$$ O₂(g)$$\rightarrow$$ 4CO₂(g) + 5H₂O(l)
$$\Delta$$H degrees Celsius = -2877 kJ

I am not sure at all how to do this and this is something we went over very briefly in class without any examples, so any help would be greatly appreciated.

 

[queenofbabes]

You were given the heat of enthalpy of...?

What does the heat of enthalpy mean anyway?

How much butane do you have to start with?

 

[Blueshift5]

Based on the given equation, the combustion of 1 mole of C4H10(g) releases 2877 kJ of heat. Therefore, to calculate the heat evolved in the complete combustion of 1.325 g C4H10(g), we first need to convert the given mass of butane to moles.

1 mole of C4H10(g) has a molar mass of 58.12 g. So, 1.325 g of C4H10(g) is equivalent to 1.325/58.12 = 0.0228 moles of C4H10(g).

Now, using the given value of -2877 kJ for the combustion of 1 mole of C4H10(g), we can calculate the heat evolved in the combustion of 0.0228 moles of C4H10(g) as follows:

Heat evolved = -2877 kJ/mol x 0.0228 mol = -65.62 kJ

Since the value of \DeltaH is negative, it indicates that the reaction is exothermic and releases heat. Therefore, the complete combustion of 1.325 g C4H10(g) at 23.6 degrees Celsius and 738 mmHg will release 65.62 kJ of heat.