Question

What mass of propane (C3H8) must be burned in a barbecue grill to release 4560 kJ of heat?

Answer 1

Answer:

The answer is "$\bold{90.62 \ g \ C_3H_8}$".

Explanation:

Write a balanced thermochemical formula for both the combustion with one mole of $C_3H_8$

$C_3H_8\ (g) +5 O_2 \ (g)\longrightarrow 3 CO_2 \ (g) +4H_2O \ (l)\ \ \ \ \ \ \ \ \Delta H_{comb} = -2219 \frac{kj}{mol}$

Calculating the molar mass of $C_3H_8$ .  

$3 (12.0107 \frac{g}{mol}) +8 (1.0079 \frac{g}{mol})= 44.0953 \frac{g}{mol}$

1) Start with both the energy released.  

2) Just use molar enthalpy of $C_3H_8$  as just a balanced equation to transfer kilojoules to birthmarks of $C_3H_8$ .  

3) It use molarity of $C_3H_8$   as a balanced equation to transfer caterpillars of $C_3H_8$  to ounces of $C_3H_8$ .

$= -4560 \ kJ \ \ \ \ \frac{1 \ mol C_3H_8}{-2219 \ kJ} \ \ \ \ \frac{44.0953 \ g \ C_3H_8}{1 \ mol \ C_3H_8}\\\\ = 90.62 \ g \ C_3H_8$