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) = Cl_{2 (g)} + H_{2 (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?

Answer 1

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