Answer
91.5%
Step-by-step explanation
Given:
2 K (s) + Br2 (1)→2 KBr (s)
mass of bromide = 22.2 g
molar mass of bromide = 159.808 g/mol
mass of KBr = 30.3 g
molar mass of KBr = 119.002 g/mol
Required: %Yield of the reaction
Solution:
Step 1: Calculate the moles of Br2
n = m/M where n is the moles, m is the mass and M is the molar mass
n = 22.2g/159.808 g/mol
n = 0.139 mol
Step 2: Use the stoichiometry to find the moles of KBr
The molar ratio between Br2 and KBr is 1:2
Therefore the number of moles of KBr = 0.139 x 2 = 0.278 mol
Step 3: Now that we have the moles, we can calculate the theoretical mass
m = n x M
m = 0.278 mol x 119.002 g/mol
m = 33.083 g
Step 4: Calculate the %Yield
Actual mass collected = 30.3 g
Theoretical mass = 33.083 g
%Yield = (actual yield/theoretical yield) x 100
%Yield = (30.3g/33.083g) x 100
%Yield = 91.5 %