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Glucose, C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the equation C6H12O6(aq)+6O2(g)⟶6CO2(g)+6H2O(l)
Calculate the number of grams of oxygen required to convert 53.0 g of glucose to CO2 and H2O.

Respuesta :

Answer : The mass of oxygen required are, 56.448 grams

Explanation : Given,

Mass of glucose = 53 g

Molar mass of glucose = 180.156 g/mole

Molar mass of [tex]O_2[/tex] = 32 g/mole

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{Mass of }C_6H_{12}O_6}{\text{Molar mass of }C_6H_{12}O_6}=\frac{53.0g}{180.156g/mole}=0.294moles[/tex]

Now we have to calculate the moles of [tex]O_2[/tex].

The balanced chemical reaction will be,

[tex]C_6H_{12}O_6(aq)+6O_2(g)\rightarrow 6CO_2(g)+6H_2O(l)[/tex]

From the balanced chemical reaction, we conclude that

As, 1 mole of [tex]C_6H_{12}O_6[/tex] react with 6 moles of [tex]O_2[/tex]

So, 0.294 mole of [tex]C_6H_{12}O_6[/tex] react with [tex]0.294\times 6=1.764[/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=(1.764mole)\times (32g/mole)=56.448g[/tex]

Therefore, the mass of oxygen required are, 56.448 grams