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?

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:

2Na + 2H₂O → 2NaOH + H₂,

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.

it is stable as it is in group 0 which have full outer shells of electrons