Question

consider the following balanced reaction. if you begin with 10.00 grams of o₂ and 25.00 grams of h₂, which is the limiting reactant and how much h₂o will be formed?
2h₂ + o₂ → 2h₂o
o h₂, 28.13 g h₂o
o h₂, 112.5 g h₂o
o o₂, 112.5 g h₂o
o o₂, 28.13 g h₂o

Explanation:

Step1: Calculate moles of \(O_2\) and \(H_2\)

Molar mass of \(O_2 = 32\space g/mol\), moles of \(O_2=\frac{10.00\space g}{32\space g/mol}= 0.3125\space mol\)
Molar mass of \(H_2 = 2\space g/mol\), moles of \(H_2=\frac{25.00\space g}{2\space g/mol}=12.5\space mol\)

Step2: Determine limiting reactant

From the balanced equation \(2H_2 + O_2
ightarrow2H_2O\), mole ratio of \(H_2:O_2 = 2:1\)
For \(0.3125\space mol\) of \(O_2\), moles of \(H_2\) required \(=2\times0.3125 = 0.625\space mol\)
Since \(12.5\space mol\) of \(H_2\) is available (\(12.5>0.625\)), \(O_2\) is the limiting reactant.

Step3: Calculate moles of \(H_2O\) formed

From the balanced equation, mole ratio of \(O_2:H_2O = 1:2\)
Moles of \(H_2O\) formed \(=2\times0.3125 = 0.625\space mol\)

Step4: Calculate mass of \(H_2O\)

Molar mass of \(H_2O=18\space g/mol\)
Mass of \(H_2O = 0.625\space mol\times18\space g/mol = 11.25\space g\)

Answer:

\(O_2\) is the limiting reactant and \(11.25\space g\) of \(H_2O\) is formed.