Question

Glucose, C6H12O6, is metabolized in living systems according to the reaction C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(l)

How many grams of water can be produced from the reaction of 24.5 g of glucose and 6.30 L of O2 at 1.00 atm and 37 ∘C?

Answer 1

our reaction is C6H12O6 + 6O2-------->6CO2 + 6H2O

molecular weight of glucose is 180g.

molecular weight of O2 is 32g

obsreving the reaction . when 1 mole of glucose is react with 6 mole of oxygen then 6 mole of water is formed

use ideal gas equation for O2= PV=nRT

p=1 atm

v=6.3L

t=37+273=310k

now n of O2 = (1*6.3)/(.08205*310)=0.2476 mol

now n of glucose = 24.5/180=0.136 mol

now we see that 0.136 mole gucose required 0.816 mole of O2 according to given reaction.

so here oxygen is limiting reagent.

now 0.2476 mol of oxygen react with .2476/6 mole of glucose . that is 0.04142 mole of glucose .

now again observe the reaction O2 and H2O have same no of mole . so mole of water is 0.2476

now mass of water = 0.2476*18= 4.4568.

molecular mass of water is 18