Limitations on reaction field

Question

Determine the limiting reactant, the mass of water formed (in g), and the mass that remains of the excess reactant when 20.00 g of HCl reacts with 20.0 g of water.

4CHl (g) + O2 (g) —> 2CL2 (g) + 2H2O (g)

Answer 1

Jordan,

One easy way to find the limiting reactant is to divide the moles of each reactant by its coefficient in the balanced equation and whichever value is lower will be the limiting reactant.

I've re-written your equation to what I believe it was meant to be...

4HCl (g) + O₂ (g) —> 2Cl₂ (g) + 2H₂O (g) ... balanced equation

For HCl: 20.00 g HCl x 1 mol / 36.5 g = 0.548 mols HCl (÷4 -> 0.137)

For O2: 20.0 g O2 x 1 mol / 32 g = 0.625 mols O2 (÷1 -> 0.625)

HCl is limiting and O₂ is in excess

Mass H₂O formed = 0.548 mols HCl x 2 mols H₂O / 4 mols H₂O x 18 g H2O/mol = 4.93 g H₂O

Mass of O₂ remaining = original mass O₂ - mass of O₂ used up

Mass of O₂ used = 0.548 mol HCl x 1 mol O2 / 4 mol HCl x 32 g O2/mol = 4.38 g O2 used

Original mass O₂ present = 20.00 g

Mass of O₂ remaining = 20.0 g - 4.38 g = 15.6 g of O₂ remaining