ThermochemistryHard
Question
The heat evolved in the combustion of glucose C6H12O6 is −680 kcal/mol. The mass of CO2 produced, when 170 kcal of heat is evolved in the combustion of glucose is
Options
A.45 g
B.66 g
C.11 g
D.44 g
Solution
$C_{6}H_{12}O_{6}(s) + 6O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(l); \Delta H = - 680\text{ Kcal}$
For 680 Kcal, 6 × 44 gm CO2 is produced. Hence, for 170 Kcal, mass of CO2 produced $= \frac{6 \times 44}{680} \times 170 = 66\text{ gm}$
Create a free account to view solution
View Solution FreeMore Thermochemistry Questions
Butane exists in various conformations in nature. At any given instant, the probability that a given butane molecule is ...Which of the following statement(s) is/are ture?...Consider the following isomerization process.CH2=CH–CH2–CH=CH2(g) → CH2=CH–CH=CH–CH3(g)Which of the following statement(...The enthalpy of formation of HCl(g) from the following reactionH2(g) + Cl2(g) → 2HCl(g) + 44 kcal is...2MnO4 – + 16H+ + 10Cl– → 2Mn2+ + 5Cl2(g) + 8H2OThe above reaction is endothermic and hence, the actual temperature of th...