For the combustion of sucrose:

Question

C12H22O11 + 12O2 → 12CO2 + 11H2O
There are 12.0 g of sucrose and 10.0 g of oxygen reacting in lab conditions.
How much H2O will be produced? (You will need notebook paper to solve this)

Answer 1

I don't have notebook paper but I suppose a computer will do.
This is a limiting reagent (LR) problem. You know that is the case when amounts are given for more than one reactant in an equation.
C12H22O11 + 12O2 → 12CO2 + 11H2O
mols sucrose = grams/molar mass = 12.0/342 = approx 0.035
How much H2O would this produce? That's 0.035 mols C12H22O11 x (11 mols H2O/1 mol C12H22O11) = 0.035 x 11 = approx 0.38 mols H2O.

mols O2 = 10.0/32 = 0.31
How much H2O would the O2 produce? That's 0.31 x (11 mols H2O/12 mol O2) = 0.28.
The smaller number is provided by O2; therefore, O2 is the LR.
grams H2O = mols H2O x molar mass H2O = 0.28 x 18 = ?
Check my arithmetic. Note that I've estimated all of the numbers. You should recalculate all of the steps and carry it out to 3 places. Post your work if you get stuck.