Question #d0f0e

1. Find the molar mass of the involved compounds which are obtainable from the periodic table.

#C_6H_12O_6=(180g)/(mol)#

#H_2O=(18g)/(mol)#

1. Given the mass of the glucose, per convention, convert the #"gram"# to #"mole"#; i.e;

#=32cancel(gC_6H_12O_6)xx(1molC_6H_12O_6)/(180cancel(gC_6H_12O_6))#

#=0.1778molC_6H_12O_6#

1. Referring to the given balanced equation for the mole ratio, find the #etaH_2O# through molar conversion; i.e;

#=0.1778cancel(molC_6H_12O_6)xx(6molH_2O)/(1cancel(molC_6H_12O_6))#

#=1.0667molH_2O#

1. Now, find the #mH_2O#.

#=1.0667cancel(molH_2O)xx(18gH_2O)/(1cancel(molH_2O))#

#=19.2gH_2O#

We have the formula;

#C_6H_12O_6 + 6O_2 -> 6CO_2 + 6H_2O#

Mole ratio

#1 : 6 -> 6 : 6#

Hence #1# mole of glucose gives #6# moles of water

Mass ratio

#180g : 192g -> 264g : 108g#

It then means that;

#180g# will yield #108g# of water

#32g# will give #x# of water

#rArr 180/32 = 108/x#

Cross multiplying..

#rArr 180 xx x = 108 xx 32#

#rArr 180x = 3456#

Divide both sides by #180#

#rArr (180x)/180 = 3456/180#

#rArr (cancel180x)/cancel180 = 3456/180#

#rArr x = 3456/180#

#rArr x = 19.2g#