2na(s) + 2h2o(l) → 2naoh(aq) + h2(g) when 2 moles of na react with water at 25°c and 1 atm, the volume of h2 formed is 24.5 l. calculate the magnitude of work done in joules when 0.45 g of na reacts with water under the same conditions. (the conversion factor is 1 l · atm = 101.3 j.)

Magnitude of work done = 24.28 J

Explanation:

No. of moles = Reacting mass/ Molar mass

Reacting mass of Na = 0.45 g

Molar mass of sodium = 23g/mol

∴ No. of moles of sodium = 0.45/23 = 0.0196 mole

If 2 moles of Na react with water at 25°C and 1 atm, 24.5 L of H₂ was formed.

∴ when 0.0196 mole of Na react with water under the same conditions, (24.5×0.0196)/2 L of H₂  will be formed.

⇒ (24.5×0.0196)/2 L = 0.24 L

0.24 L × 1 atm = 0.24 L . atm

Since  1 L · atm = 101.3 J

∴      0.24 L . atm = (0.24 L . atm ×101.3)/1 = 24.28 J

Magnitude of work done = 24.28 J

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l