Identify the limiting reactant when 1.22 g of o2 reacts with 1.05 g h2 to produce water.

The reaction between oxygen, O2, and hydrogen, H2, to produce water can be expressed as,

                    2H2 + O2 --> 2H2O

The masses of each of the reactants are calculated below.

          2H2 = 4(1.01 g) = 4.04 g
          O2 = 2(16 g) = 32 g

Given 1.22 grams of oxygen, we determine the mass of hydrogen needed.
        (1.22 g O2)(4.04 g H2 / 32 g O2) = 0.154 g of O2

Since there are 1.05 grams of O2 then, the limiting reactant is 1.22 grams of oxygen.

1.22 g of oxygen

29.1-12=17.1the answer is this one