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 25 °C and a pressure of 759 mm Hg. If the wet H2 gas formed occupies a volume of 8.11 L, the number of grams of H2 formed is _______g. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

0.648 g

Step-by-step explanation:

Step 1: Write the balanced equation

2 Na(s) + 2 H₂O(l) ⇒ 2 NaOH(aq) + H₂(g)

Step 2: Calculate the partial pressure of H₂

The total pressure (P) is equal to the sum of the partial pressures of hydrogen and water vapor.

P = pH₂O + pH₂

pH₂ = P - pH₂O

pH₂ = 759 mmHg - 23.8 mmHg = 735 mmHg

Step 3: Convert "pH₂" to atm

We will use the conversion factor 1 atm = 760 mmHg.

735 mmHg × 1 atm/760 mmHg = 0.967 atm

Step 4: Convert "25°C" to K

We will use the following expression.

K = °C + 273.15 = 25°C + 273.15 = 298 K

Step 5: Calculate the moles of hydrogen (n) formed

We will use the ideal gas equation.

P × V = n × R × T

n = P × V/R × T

n = 0.967 atm × 8.11 L/(0.0821 atm.L/mol.K) × 298 K = 0.321 mol

Step 6: Calculate the mass corresponding to 0.321 moles of hydrogen

The molar mass of H₂ is 2.02 g/mol.

0.321 mol × 2.02 g/mol = 0.648 g