Consider the synthesis of water as shown in Model 3. A container is filled with 10.0 g of H2 and 5.0 g of O2. a. Which reactant (hydrogen or oxygen) is the limiting reactant in this case? Show your work. Hint: Notice that you are given reactant quantities in mass units here, not moles.
b. What mass of water can be produced? Show your work.
c. Which reactant is present in excess, and what mass of that reactant remains after the reaction is complete? Show your work.

2H2 + O2 → 2H2O

First, we will find the moles :

n H2 = m H2 / Mr H2

n H2 = 10 / 2

n H2 = 5 mole

n O2 = m O2 / Mr O2

n O2 = 5 / 16

n O2 = 0.3125 mole

n H2 / coef. H2 > n O2 / coef. O2

So, O2 is the limiting reactant

The mass of water produced :

n H2O = (coef. H2O / coef. O2) • n O2

n H2O = (2/1) • 0.3125

n H2O = 0.625 mole

m H2O = n H2O • Mr H2O

m H2O = 0.625 • 18

m H2O = 11.25 gr

Excesses reactant :

n H2 = 5 - 0.625 = 4.375 mole

m H2 = n H2 • Mr H2

m H2 = 4.375 • 2

m H2 = 8.75 gr

a graduated cylinder.

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