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.

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.