When 180 g of glucose is subjected to combustion, the volume of $C O_{2}$ liberated at STP is :

22.4 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

67.2 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

44 L

No worries! We‘ve got your back. Try BYJU‘S free classes today!

134.4 L

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is A 134.4 L
The balanced chemical equation for combustion of glucose can be written as:
$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$
$180$ g of $C_{6} H_{12} O_{6}$ produces $22.4 \times 6$ $= 134.4$ L of $C O_{2}$

Suggest Corrections

Similar questions

C O_{2}

C O_{2}

C O_{2}

C O_{2}

Join BYJU'S Learning Program