Here is your balanced equation:

#C_6H_12O_6+ 6O_2rarr 6H_2O+6C0_2#

We start Dimensional analysis (the factor-label method) by writing the given information over 1:

#(35"g of "CO_2)/1#

We look up the molar mass of #CO_2# and multiply by that factor so that it converts the units to moles:

#(35"g of "CO_2)/1(1"mol of "CO_2)/(44.01"g of "CO_2)#

Please observe how the units cancel and we are left with moles of #CO_2#:

#(35color(red)(cancel("g of "CO_2)))/1(1"mol of "CO_2)/(44.01color(red)(cancel("g of "CO_2)))#

We obtain the factor that 6 moles of water are produced when 6 moles of #CO_2# are produced, from the equation:

#(35color(red)(cancel("g of "CO_2)))/1(1color(red)cancel(("mol of "CO_2)))/(44.01color(red)(cancel("g of "CO_2)))(6"mol of "H_2O)/(6color(red)(cancel("mol of "CO_2)))#

Look up the molar mass of water:

#(35color(red)(cancel("g of "CO_2)))/1(1color(red)cancel(("mol of "CO_2)))/(44.01color(red)(cancel("g of "CO_2)))(6color(red)cancel(("mol of "H_2O)))/(6color(red)(cancel("mol of "CO_2)))(18.02"g of "H_2O)/(1color(red)cancel(("mol of "H_2O)))#

We have the units in grams of #H_2O#; now, just do the multiplication or division:

#35/44.01(18.02)= 14.33"g of "H_2O#