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.
<em>2Na + 2H₂O → 2NaOH + H₂,</em>
- The balanced equation shows that <em>2.0 moles of Na metal </em>react with 2.0 moles of water to produce 2.0 moles of NaOH and <em>1.0 mole of H₂</em>,
- Firstly, we need to convert the volume of H₂ (7.80 L) produced to no. of moles (n) using the ideal gas law: <em>PV = nRT</em>,
where, P is the pressure of the gas in atm<em> (P at STP = 1.0 atm)</em>,
V is the volume of the gas in L <em>(V = 7.80 L)</em>,
n is the number of moles in mole,
R is the general gas constant<em> (R = 0.082 L.atm/mol)</em>,
T is the temperature of the gas in K <em>(T at STP = 0.0 °C + 273 = 273.0 K)</em>.
∴ 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.
<em>Using cross multiplication:</em>
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.
