Question

Given that 16.0 g of MnO2 and 30.0 g of HCl react according to
MnO2 + 4HCl -------> MnCl2 + Cl2 + 2H2O
What is limiting reagant?
What mass of MnCl2 could be produced?

Answer 1

16.0g MnO2 x( molecular mass MnCl2/molecular mass of MnO2)= mass of MnCl2 (1)

30.0 g HCl x (molecular mass MnCl2/molecular mass of HCl)= mass of MnCl2 (2)

(1) < (2) then MnO2 = limiting reactant, and (1) is the theoretical yield or mass MnCl2
(1) > (2) then HCl = limiting reactant, and (2) is the theoretical yield or mass MnCl2

Answer 2

so how much of each reagant remains when the reaction is complete?

Answer 3

and assuming a % yield of 71.2%. calculate the actual yield of MnCl2

Answer 4

@Cuanchi thanks!!

Answer 5

Hold down I forgot to include the conversion factor for the HCl equation the stoichiometry is 4moles HCl/1mole MnCl2

30.0 g HCl x (molecular mass MnCl2/4 molecular mass of HCl)= mass of MnCl2 (2)

Answer 6

actual yield= 71.2 x Theoretical yield/100

Answer 7

its like when you go to the store and said 25% OFF, how much do you pay? this is the actual yield