Question

when 180 grams of glucose is subjected to combustion, the volume of CO2 liberated at STP is ??

Answer 1

Answer:

6CO2+6O2 treated by sunlight C6H12O6+ 6O2

Answer 2

Explanation:

The balanced chemical equation for combustion of glucose can be written as:

C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}OC6H12O6+6O2→6CO2+6H2O

180180 g of C_{6}H_{12}O_{6}C6H12O6 produces 22.4\times 622.4×6 = 134.4=134.4 L of CO_{2}CO2