Calculate the heat released when 5.00 l of cl2(g) with a density of 1.88 g/l reacts with an excess of sodium metal at 25°c and 1 atm to form sodium chloride.

Answer : The heat release in the formation of NaCl will be -108.8 KJ

Explanation : We are given 5 L of  which has density of 1.88 g

So, 5 X 1.88 = 9.4 g;

Now, 9.4 g of  = 9.4 / 71 mole = 0.134 moles of  

From the reaction we know 2 X 0.134 = 0.264 moles of NaCl is formed.

Now, Heat of formation of NaCl is - 411 KJ;

So, the heat release for formation of   = 0.264 X (-411)
 = - 108.8 KJ