Question

The combustion of propane may be described by the chemical equation C 3 H 8 ( g ) + 5 O 2 ( g ) ⟶ 3 CO 2 ( g ) + 4 H 2 O ( g ) How many grams of O 2 ( g ) are needed to completely burn 89.2 g C 3 H 8 ( g ) ?

Answer 1

Answer: The mass of $O_2$ needed is, 13.59 grams.

Explanation : Given,

Mass of $C_3H_8$ = 89.2 g

Molar mass of $C_3H_8$ = 42 g/mol

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $C_3H_8$

$\text{Moles of }C_3H_8=\frac{\text{Given mass }C_3H_8}{\text{Molar mass }C_3H_8}$

$\text{Moles of }C_3H_8=\frac{89.2g}{42g/mol}=2.124mol$

Now we have to calculate the moles of $C_3H_8$

The balanced chemical equation is:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)$

From the reaction, we conclude that

As, 1 mole of $C_3H_8$ react with 5 mole of $O_2$

So, 2.124 mole of $C_3H_8$ react with $\frac{2.124}{5}=0.4248$ mole of $O_2$

Now we have to calculate the mass of $O_2$

$\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2$

$\text{ Mass of }O_2=(0.4248moles)\times (32g/mole)=13.59g$

Therefore, the mass of $O_2$ needed is, 13.59 grams.