% Yield, %Purity, and % Error
%Yield =
got in lab
100
calc. should get
%Purity =
got in lab
100
Calc should get
% Error =100 - % Yield
2 HMnO4(aq) + 5 Na3AsO3(s) + 19 HNO3(l)  2 Mn(NO3)2(s) + 5 H3AsO4(l) + 15 NaNO3(aq) + 3 H2O(g)

(192)
(63)
(179)
(142)
(85)
(18)
Density=1.35g/mL
Density=1.7g/mL
**This problem is continued from #2 on the Limiting and Excess Reactants Worksheet**
1. The lab group working at table B mixes 450 grams HMnO4 with 4 x 1025 atoms Na3 AsO3 and 1.50
liters HNO3. They form 2,100 g NaNO3 in the experiment. Find %Yield, % Purity, % Error.
From before HNO3 is the limiting reactant and we calculated 2,157 g NaNO3 produced.
molar masses (120)
(g/mol)
got in lab
2,100g
100 
100  97.4% = 97.4% Purity and subtracting
calc. should get
2,157g
from 100-97.4= 2.6% Error
Therefore %Yield =
2. We react 300 g HMnO4 AND 4.5 x 1024 molecules Na3AsO3 AND 1,000 mL HNO3 and form 70
liters H2O gas in the lab. Find % Yield
3. If we mix 30 grams HMnO4 with excess Na3AsO3 and 115 mL HNO3 and produce 47.5 mL H3AsO3
in the experiment, then find percent error.
4. Mix excess HMnO4, 2 x 1023 molecules Na3AsO3 and 70 mL HNO3. Lab group F makes 7 x 1022
molecules Mn(NO3)2 in the lab, find % Purity. Who would this be % Purity of?
5. GRAND FINALE
We react 325 grams HMnO4 AND 4.2 x 1025 atoms Na3AsO3 AND 1,350 mL HNO3 and form 1,650 g
NaNO3 in the lab. Find %Yield, %Purity, % Error and the amount of each leftover reactant.