A chemical plant uses electrical energy to produce gaseous Cl 2 and H 2 and aque

Question

A chemical plant uses electrical energy to produce gaseous Cl2 and H2 and aqueous NaOH from aqueous sodium chloride and water. Cl2 and NaOH are important feedstocks for producing other chemicals, while H2 is an important fuel and chemical feedstock. Consider that you are starting out with 1.5 x 105 kg of NaCl and 1.5 x 105 kg of H2O.

2 NaCl (aq) + 2 H2O (l) = Cl2 (g) + H2 (g) + 2 NaOH (aq)

A. What is the limiting reactant?

B. What is the theoretical yeild of Cl2 in grams?

C. What is the excess reactant and how many grams of excess reactant remain?

D If the actual yield of Cl2 for the given reaction conditions is 1.17 x 105 kg, what is the percent yield?

Explanation / Answer

mol NaCl = mass/MW = (1.5*10^5)/58 = 2586.20 mol of NaCl

mol of H2O = mass/MW = (1.5*10^5)/(18) = 8333.33 mol of H2O

ratio is 2:2, therefore 1:1; then we have excess of H2O

a)

limiting reactant is NaCl, since there is low amount of mol

b)

yield of Cl2 in grams

ratio is 2:1 therefore

2586.20 mol of NaCl = 2586.20 /2 Cl = 1293.1 mol of Cl2

mass = mol*MW = 1293.1*70 = 90517 kg of Cl2

c)

excess reactant is H2O

grams excess:

8333.33 -2586.20 = 5747.13 kmol of H2O in excess

mass = mol*MW = 5747.13*18 = 103448.34 kg of H2O are left

d)

actualy yield if m = 1.17*10^5 is achieved

yield = (real/theoreticla)*100 = (1.17*10^5)/90517 *100 = 129.25%

which is impossible, since you ar eproducing plenty of Cl2, there is no possible way to do that

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