Combustion of glucose (C_6H_{12}O_6) produces | vedclass

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(B) The balanced chemical equation for the combustion of glucose is:$C_6H_{12}O_{6(s)} + 6O_{2(g)} \longrightarrow 6CO_{2(g)} + 6H_2O_{(\ell)}$First,c

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$960$

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(B) The balanced chemical equation for the combustion of glucose is:$C_6H_{12}O_{6(s)} + 6O_{2(g)} \longrightarrow 6CO_{2(g)} + 6H_2O_{(\ell)}$First,calculate the number of moles of glucose:$	ext{Moles of glucose} = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{900 \ g}{180 \ g \ mol^{-1}} = 5 \ mol$From the stoichiometry of the reaction,$1 \ mol$ of glucose requires $6 \ mol$ of $O_2$.Therefore,$5 \ mol$ of glucose requires $5 	imes 6 = 30 \ mol$ of $O_2$.Finally,calculate the mass of $O_2$ required:$	ext{Mass of } O_2 = 	ext{Moles} 	imes 	ext{Molar mass of } O_2 = 30 \ mol 	imes 32 \ g \ mol^{-1} = 960 \ g$

Combustion of glucose $(C_6H_{12}O_6)$ produces $CO_2$ and water. The amount of oxygen (in $g$) required for the complete combustion of $900 \ g$ of glucose is:
[Molar mass of glucose in $g \ mol^{-1} = 180$]

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Combustion of glucose $(C_6H_{12}O_6)$ produces $CO_2$ and water. The amount of oxygen (in $g$) required for the complete combustion of $900 \ g$ of glucose is:[Molar mass of glucose in $g \ mol^{-1} = 180$]