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?
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.