Question

Consider the combustion reaction of propane: C3H8(g) + 5 O2(g) ® 3 CO2(g) + 4 H2O(g), where DH = – 531 kcal/mol. If 6.70 ´ 104
kcal of energy is released in the reaction, how many grams of oxygen were consumed?

Answer 1

moles of propane =

6.70 x 10^4 kcal/531 kcal/mole

moles of O2 per the balanced equation  = 5* moles of propane

mass of O2 = moles of O2 * 32 g/mole . 20.2 kg

hope this helps

Answer 2

Answer:- 20200 g of oxygen are consumed.

Solution:- It asks to caculate the grams of oxygen consumed when 67000 kcal of energy is released.

The given balanced equation is:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)\Delta H=-531\frac{kcal}{mol}$

From balanced equation, 531 kcal of energy is released when 5 moles of oxygen are consumed. We can easily calculate the moles of oxygen consumed when 67000 kcal of energy is released and the moles on multiplying by molar mass are converted to grams.

the calculations are shown using dimensional analysis.

Molar mass of oxygen is 32 g per mol.

$6.70*10^4kcal(\frac{5molO_2}{531kcal})(\frac{32gO_2}{1molO_2})$

= $20188gO_2$

If we round the number to three sig figs then it becomes 20200 g.

So, 20200 g of $O_2$ are consumed.