Respuesta :
Answer : The mass of [tex]O_2[/tex] required is, 61.82 grams.
Explanation :
First we have to calculate the moles of [tex]C_6H_{12}O_6[/tex]
[tex]\text{Moles of }C_6H_{12}O_6=\frac{\text{Given mass }C_6H_{12}O_6}{\text{Molar mass }C_6H_{12}O_6}=\frac{58.0g}{180g/mol}=0.322mol[/tex]
Now we have to calculate the moles of [tex]O_2[/tex].
The balanced chemical equation is:
[tex]C_6H_{12}O_6(aq)+6O_2(g)\rightarrow 6H_2O(l)+6CO_2(g)[/tex]
From the balanced reaction we conclude that
As, 1 mole of [tex]C_6H_{12}O_6[/tex] react with 6 mole of [tex]O_2[/tex]
So, 0.322 moles of [tex]C_6H_{12}O_6[/tex] react with [tex]0.322\times 6=1.932[/tex] moles 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]
Molar mass of [tex]O_2[/tex] = 32 g/mole
[tex]\text{ Mass of }O_2=(1.932moles)\times (32g/mole)=61.82g[/tex]
Therefore, the mass of [tex]O_2[/tex] required is, 61.82 grams.
