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

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.