Question

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.

Answer 1

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