If you burn 32.3g of hydrogen and produce 288g of water, how much oxygen reacted?

256 g  

Step-by-step explanation:

a) Balanced equation

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:    2.016   32.00    18.02

            2H₂   +   O₂ ⟶ 2H₂O

m/g:     32.3                    288

(b) Moles of H₂O

n = 288 g H₂O × (1 mol H₂O/18.02 g H₂O)

  = 15.98 mol H₂O

(c) Moles of O₂

The molar ratio is (2 mol H₂O /1 mol O₂)

n = 15.98 mol H₂O × (1 mol O₂ /2 mol H₂O)

  = 7.991 mol O₂

(d) Mass of O₂

m = 7.991 mol O₂× (32.00 g O₂/1 mol O₂)

    = 256 g O₂

The reaction requires 256 g O₂.

Note: If you base your answer on the mass of hydrogen, you get the same result.

Mass of O₂ = 32.3 g H₂ × 1 mol H₂/2.016 g H₂

                  = 16.02 mol H₂ × 1 mol O₂ /2 mol H₂

                  = 8.011 mol O₂ × 32.00 g O₂/1 mol O₂  

                  = 256 g O₂

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.