Calculate the work done by the expansion of hydrogen gas when 7.1 g of sodium react with water at constant temperature of 27°C and pressure of 1.7 atm. Use a precision of 2 significant digits. 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g)

-390J

Explanation:

The given reaction is,

2Na(s) + 2H2O —-> 2NaOH + H2

Number of moles of NaOH = Mass/molar mass

= 7.1/23 = 0.31 mol

Since, 2 mol Na produces 1 mol H2

Hence, 0.31 mol Na will produce

= 0.31 mol /2 = 0.155 mol H2

According to ideal gas equation,

PV = n RT

V = nRT/P

= 0.155mol x 0.0821 L atm K-1 mol-1 x (27+273 K) / 1.7atm

= 2.25L

Therefore,

Work of expansion = - PV = - 1.7 atm x 2.25L

= - 3.825atm L

= - 3.825 x 101.325 J = - 387.57 J

= - 390 J

one negative thing it that it can kill water animals. another negative thing is it can cause the destruction of ecosystem. one way water pollution can be reduced is to conserve energy.