Calculate the value of work w for the following system if 26

Calculate the value of work, w, for the following system if 26.5 g of NaN3 reacts completely at 1.00 atm and 295K.

2NaN3(s) ----> 2Na(s) + 3N2(g)

Work done (w)

= nRT

For the given reaction,

n = moles of NaN3

R = gas constant

T = 295 K

we get,

work done = 26.5 x 8.314 x 295/65

= 999.92 J