In the process of respiration in plants,180 , | vedclass

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(C) The balanced chemical equation for aerobic respiration is:$C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + 674 \, kcal$Calculating the molecular w

Vedclass · Accepted Answer

$264 \, g$ of $CO_2$,$108 \, g$ of water and $674 \, kcal$ of energy

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(C) The balanced chemical equation for aerobic respiration is:$C_6H_{12}O_6 + 6O_2 ightarrow 6CO_2 + 6H_2O + 674 \, kcal$Calculating the molecular weights:$1$ mole of glucose $(C_6H_{12}O_6)$ = $(6 	imes 12) + (12 	imes 1) + (6 	imes 16) = 72 + 12 + 96 = 180 \, g$.$6$ moles of oxygen $(6O_2)$ = $6 	imes (2 	imes 16) = 6 	imes 32 = 192 \, g$.Calculating the products:$6$ moles of carbon dioxide $(6CO_2)$ = $6 	imes (12 + 32) = 6 	imes 44 = 264 \, g$.$6$ moles of water $(6H_2O)$ = $6 	imes (2 + 16) = 6 	imes 18 = 108 \, g$.Energy released = $674 \, kcal$.Therefore,$180 \, g$ of sugar and $192 \, g$ of oxygen produce $264 \, g$ of $CO_2$,$108 \, g$ of water,and $674 \, kcal$ of energy.

In the process of respiration in plants,$180 \, g$ of sugar plus $192 \, g$ of oxygen produce:

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