Question

how many moles of oxygen would be consumed to produce 68.1 g water? c₃h₈ + 5o₂ → 3co₂ + 4h₂o

Explanation:

Step1: Calculate moles of water

The molar mass of \(H_2O\) is \(M=(2\times1 + 16)\space g/mol=18\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 68.1\space g\) and \(M = 18\space g/mol\), we have \(n_{H_2O}=\frac{68.1\space g}{18\space g/mol}\approx3.783\space mol\).

Step2: Use stoichiometry from the balanced equation

From the balanced equation \(C_3H_8 + 5O_2
ightarrow3CO_2 + 4H_2O\), the mole ratio of \(O_2\) to \(H_2O\) is \(\frac{n_{O_2}}{n_{H_2O}}=\frac{5}{4}\).
So \(n_{O_2}=\frac{5}{4}\times n_{H_2O}\).
Substitute \(n_{H_2O} = 3.783\space mol\) into the equation: \(n_{O_2}=\frac{5}{4}\times3.783\space mol\approx4.73\space mol\).

Answer:

\(4.73\space mol\)