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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.

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

Explanation : We are given 5 L of [tex] Cl_{2} [/tex] which has density of 1.88 g

So, 5 X 1.88 = 9.4 g;

Now, 9.4 g of [tex] Cl_{2} [/tex] = 9.4 / 71 mole = 0.134 moles of [tex] Cl_{2} [/tex] 

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 [tex] Cl_{2} [/tex]  = 0.264 X (-411)
 = - 108.8 KJ

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