Question

sodium metal reacts with water to produce hydrogen gas according to the following equation:
2na(s) + 2h₂o(l) → 2naoh(aq) + h₂(g)
the product gas, h₂, is collected over water at a temperature of 20°c and a pressure of 747.0 mm hg. if the wet h₂ gas formed occupies a volume of 8.92 l, the number of moles of na reacted was
mol. the vapor pressure of water is 17.5 mm hg at 20°c.

Explanation:

Step1: Calculate the pressure of dry \(H_2\)

The total pressure \(P_{total}=747.0\ mmHg\) and the vapor pressure of water \(P_{H_2O}=17.5\ mmHg\).
Using Dalton's law of partial pressures \(P_{H_2}=P_{total}-P_{H_2O}\)
\(P_{H_2}=747.0 - 17.5=729.5\ mmHg\)
Convert pressure to atm: \(P=\frac{729.5}{760}\ atm\approx0.96\ atm\)
Convert temperature to Kelvin: \(T = 20+273=293\ K\), \(V = 8.92\ L\)

Step2: Use the ideal gas law \(PV=nRT\)

\(R = 0.0821\ L\cdot atm/(mol\cdot K)\)
From \(PV=nRT\), we can solve for \(n_{H_2}\): \(n_{H_2}=\frac{PV}{RT}\)
Substitute the values: \(n_{H_2}=\frac{0.96\times8.92}{0.0821\times293}\)
\(n_{H_2}=\frac{8.5632}{24.0553}\approx0.356\ mol\)

Step3: Use the stoichiometry of the reaction

The balanced chemical equation is \(2Na(s)+2H_2O(l)\to2NaOH(aq)+H_2(g)\)
The mole ratio of \(Na\) to \(H_2\) is \(2:1\)
So \(n_{Na}=2n_{H_2}\)
\(n_{Na}=2\times0.356 = 0.712\ mol\)

Answer:

\(0.712\)