Chlorine gas can be prepared according to the reaction: 16 HCl+2KMnO4->2MnCl2+8H2O what is the percent yield of Cl2 if 150g HCl are reacted, producing 75g Cl2?

Answer 1

To find the percent yield of Cl2, we need to calculate the theoretical yield and compare it to the actual yield.

First, let's calculate the molar mass of HCl:

HCl = 1 x 1.01 g/mol (atomic mass of hydrogen) + 1 x 35.45 g/mol (atomic mass of chlorine)
HCl = 36.46 g/mol

Next, we calculate the number of moles of HCl:

150 g HCl / 36.46 g/mol = 4.1139 mol HCl

According to the balanced equation, the molar ratio between HCl and Cl2 is 16:1. So, the theoretical yield of Cl2 can be calculated by multiplying the number of moles of HCl by the stoichiometric coefficient:

4.1139 mol HCl x (1 mol Cl2 / 16 mol HCl) = 0.2571 mol Cl2

Now, let's calculate the theoretical yield of Cl2 in grams:

0.2571 mol Cl2 x 70.91 g/mol (atomic mass of Cl2) = 18.2449 g Cl2

Finally, we can calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100:

Percent yield = (75 g Cl2 / 18.2449 g Cl2) x 100 = 411.7% (rounded to the nearest tenth)

The percent yield of Cl2 is 411.7%. Note that a percent yield above 100% indicates that there might be some experimental errors, such as incomplete reaction or impurities in the reactants.