Question in the description, can you please explain how you got the answer, thank you

Question

What is the greatest amount of water in grams that can be produced 215 grams of heptane (C₇H₁₆) and 87 grams of oxygen gas? Which reactant is limiting? How many grams of the excess reactant will be left after the reaction?

C₇H₁₆  + 11 O₂  7 CO₂ + 8 H₂O

Answer 1

Combustion of heptane:

C₇H₁₆ + 11 O₂ ==> 7CO₂ + 8H₂O ... balanced equation

215 g.........87 g....................? g........

To find the limiting reactant, we determine how many moles of H₂O can be formed from each reactant:

215 g C₇H₁₆ x 1 mol C₇H_16/100 g = 2.15 mol x 8 mol H₂O/mol C₇H₁₆ = 17.2 mol H₂O x 18 g/mol = 310 g

87 g O₂ x 1 mol O₂/32 g = 2.72 mole O₂ x 8 mol H₂O/11 mol O₂ = 1.98 mol H₂O x 18 g/mol = 35.6 g H₂O

So, O₂ is limiting and 35.5 g of H₂O will be produced

Amount of excess reactant (C₇H₁₆) left over after the reaction:

2.72 mol O₂ x 1 mol C₇H₁₆/11 mole O₂ = 0.247 moles C₇H₁₆ used up

moles left = starting moles - moles used up = 2.15 moles - 0.247 moles = 1.90 moles left over

mass left over = 1.90 moles x 100 g/mol = 190 g C₇H₁₆ left over