Answer :
[tex]C_6H_{12}O_6 + 6 \ O_2 \to 6 \ CO_2 + 6 \ H_2O[/tex]
1 mole of glucose : 6 moles of oxygen
First calculate the number of moles of oxygen in 10 g:
[tex]M=32 \ \frac{g}{mol} \\ m=10 \ g \\ n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol[/tex]
1 mole of glucose reacts with 6 moles of oxygen
x moles of glucose reacts with 0.3125 moles of oxygen
[tex]x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol[/tex]
Now calculate the mass of 5/96 moles of glucose.
[tex]M=180 \ \frac{g}{mol} \\ n=\frac{5}{96} \ mol \\ m=180 \ \frac{g}{mol} \cdot \frac{5}{96} \ mol=\frac{900}{96} \ g=9.375 \ g[/tex]
The maximum mass of glucose that can be burned in 10 g of oxygen is 9.375 g.
1 mole of glucose : 6 moles of oxygen
First calculate the number of moles of oxygen in 10 g:
[tex]M=32 \ \frac{g}{mol} \\ m=10 \ g \\ n=\frac{10 \ g}{32 \ \frac{g}{mol}}=0.3125 \ mol[/tex]
1 mole of glucose reacts with 6 moles of oxygen
x moles of glucose reacts with 0.3125 moles of oxygen
[tex]x=\frac{1 \ mol \cdot 0.3125 \ mol}{6 \ mol} = \frac{3125}{60000} \ mol= \frac{5}{96} \ mol[/tex]
Now calculate the mass of 5/96 moles of glucose.
[tex]M=180 \ \frac{g}{mol} \\ n=\frac{5}{96} \ mol \\ m=180 \ \frac{g}{mol} \cdot \frac{5}{96} \ mol=\frac{900}{96} \ g=9.375 \ g[/tex]
The maximum mass of glucose that can be burned in 10 g of oxygen is 9.375 g.
