Question

If 2.00 g hydrogen gas is reacted with 28.0 g of oxygen gas in the presence of 0.40 Pt catalyst and produces 5.00 g water. What is the limiting reagent, theoretical yield of water in moles.and percent yield? Please show all work.

Answer 1

Answer :

If 2.0 gm of H₂ reacts with 28 gm of O₂ to form 5gm of H₂O,then percent yield of the reaction,

The balanced equation is 2 H₂ + O₂ = H₂O.

one O atom for every 2 H atoms
every H atom weighs 1 unit
every H₂ molecule weighs 2 units
every O atom weighs 16 units
the sum of 2 units and 16 units is 18 units, the H₂O mass

limiting reagent is Hydrogen (H₂)

First, determine the theoretical yield. You have mass of the reactant H₂, so you need to change this to moles of H₂. Use the mole ratio and then determine the theoretical amount of H₂O, that will be made.

Mass of H₂ (use the molar mass equivalent here) = moles H₂ (use the mole ratio here to change moles to moles) = moles H₂O (use the molar mass equivalent here) = mass H₂O = percent yield.

2.0 gm H₂ x 1 mole H₂/2 gm H₂ = 1.0 moles H₂

1.0 moles H₂ x 1 mole H₂O /1 mole H₂ = 1.0 moles H₂O

1.0 moles H₂O x 18gm H₂O /1 mole H₂O = 18 gm H₂O. This is the theoretical yield of water.

Percent yield = actual yield/theoretical yield x 100%

Percent yield = 5 gm H₂O /18 gm H₂O x 100% = 27.77%