Respuesta :
Answer: The mass of [tex]O_2[/tex] needed is, 13.59 grams.
Explanation : Given,
Mass of [tex]C_3H_8[/tex] = 89.2 g
Molar mass of [tex]C_3H_8[/tex] = 42 g/mol
Molar mass of [tex]O_2[/tex] = 32 g/mol
First we have to calculate the moles of [tex]C_3H_8[/tex]
[tex]\text{Moles of }C_3H_8=\frac{\text{Given mass }C_3H_8}{\text{Molar mass }C_3H_8}[/tex]
[tex]\text{Moles of }C_3H_8=\frac{89.2g}{42g/mol}=2.124mol[/tex]
Now we have to calculate the moles of [tex]C_3H_8[/tex]
The balanced chemical equation is:
[tex]C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)[/tex]
From the reaction, we conclude that
As, 1 mole of [tex]C_3H_8[/tex] react with 5 mole of [tex]O_2[/tex]
So, 2.124 mole of [tex]C_3H_8[/tex] react with [tex]\frac{2.124}{5}=0.4248[/tex] mole of [tex]O_2[/tex]
Now we have to calculate the mass of [tex]O_2[/tex]
[tex]\text{ Mass of }O_2=\text{ Moles of }O_2\times \text{ Molar mass of }O_2[/tex]
[tex]\text{ Mass of }O_2=(0.4248moles)\times (32g/mole)=13.59g[/tex]
Therefore, the mass of [tex]O_2[/tex] needed is, 13.59 grams.
