Question

part a
consider the chemical reaction:
2h₂o(l) → 2h₂(g) + o₂(g)
what mass of h₂o is required to form 1.80 l of o₂ at a temperature of 335 k and a pressure of 0.968 atm?
express your answer in grams to three significant figures.
view available hint(s)
m = g

Explanation:

Step1: Calculate the number of moles of \(O_2\) using the ideal gas law \(PV = nRT\)

The ideal gas law is \(n=\frac{PV}{RT}\). Given \(P = 0.968\space atm\), \(V=1.80\space L\), \(R = 0.0821\space L\cdot atm/(mol\cdot K)\), \(T = 335\space K\).

$$ n=\frac{0.968\times1.80}{0.0821\times335} $$

$$ n=\frac{1.7424}{27.5035}\approx0.0634\space mol $$

Step2: Use the stoichiometry of the reaction \(2H_2O(l)\to2H_2(g)+O_2(g)\)

From the balanced equation, the mole ratio of \(H_2O\) to \(O_2\) is \(2:1\). So the number of moles of \(H_2O\), \(n_{H_2O}=2\times n_{O_2}\)
\(n_{H_2O}=2\times0.0634 = 0.1268\space mol\)

Step3: Calculate the mass of \(H_2O\) using \(m = nM\)

The molar mass of \(H_2O\), \(M=(2\times1.008 + 16.00)=18.016\space g/mol\)
\(m = 0.1268\times18.016\)

$$ m=0.1268\times18.016\approx2.28\space g $$

Answer:

\(2.28\space g\)