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

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !

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