Question

Read the given equation. 2Na + 2H2O ? 2NaOH + H2 During a laboratory experiment, a certain quantity of sodium metal reacted with water to produce sodium hydroxide and hydrogen gas. What was the initial quantity of sodium metal used if 7.80 liters of H2 gas were produced at STP?

Answer 1

Answer:

The number of moles of Na metal that used initially = 0.70 mol.

The quantity of Na metal used initially to produce 7.80 of H₂ gas = 16.02 g.

Explanation:

• It is a stichiometry problem.

2Na + 2H₂O → 2NaOH + H₂,

• The balanced equation shows that 2.0 moles of Na metal react with 2.0 moles of water to produce 2.0 moles of NaOH and 1.0 mole of H₂,

• Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: PV = nRT,

where, P is the pressure of the gas in atm (P at STP = 1.0 atm),

V is the volume of the gas in L (V = 7.80 L),

n is the number of moles in mole,

R is the general gas constant (R = 0.082 L.atm/mol),

T is the temperature of the gas in K (T at STP = 0.0 °C + 273 = 273.0 K).

∴ The number of moles of H₂ gas (n) = PV / RT = [(1.0 atm)(7.80 L)] / [(0.082 L.atm/mol.K)(273.0 K)] = 0.35 mol.

Using cross multiplication:

2.0 moles of Na will produce → 1.0 mole of H₂, from the stichiometrey.

??? moles of Na will produce → 0.35 mole of H₂.

∴ The number of moles of Na metal that used initially = (2.0 mol)(0.35 mol) / (1.0 mol) = 0.70 mol.

Now, we can get the quantity of Na metal using the relation:

∴ mass = n x molar mass = (0.70 mol)(22.989 g/mol) = 16.02 g.