Question

Pb203

+
2K3N
-)
2PBN
* 3K20
If 12.7 g of K20 were made with a percentage yield of 78.4 %, what mass of Pb20s was used?

Answer 1

Step-by-step explanation:

The Molar mass of K2O is 94,2 g/mol

The moles of K2O can be calculated

`n_(K_2 O) = (12,7 g)/(94,2 g/mol) = 0,135 mol`

The stoichiometric coefficients must be considered: in order to produce 3 moles of K2O, 1 mole of Pb2O3 are needed. In other words, if 1 mole of K2O is produced, 1/3 of a mole of Pb2O3 is needed. So

`n_(Pb2O3) = (0,135 mol)/(3) = 0,0450 mol`

The molar mass of Pb2O3 is 462,4 g/mol

The theoretical mass of Pb2O3 used to produce K2O is

`m_(Pb2O3) = 0,0450 mol * 462,4 g/mol = 20,8 g`

Since the yield is not 100%, a larger mass was needed to perform the reaction.

`(20,8)/(78,4) = (x)/(100) \\ \\ x = (20,8 * 100)/(78,4) = 26,5 g`