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Cálculos con unidades.
% Teórico

NH4NO3

100g NH₄NO₃*(1 mol NH₄NO₃)/(80,0436g NH₄NO₃)*(1 mol NH₃)/(1mol NH₄NO₃)*(17g NH₃)/(1 mol NH₃)=21,24gNH₃

NH4Cl

100g NH₄Cl*(1 mol NH₄Cl)/(53,4914g NH₄Cl)*(1 mol NH₃)/(1 mol NH₄Cl)*(17g NH₃)/(1 mol NH₃)=31,78gNH₃

(NH4)6Mo7O24.4H2O

100g sal*(1 molsal)/(1235,8594g sal)*(6 mol NH₃)/(1 mol sal)*(17g NH₃)/(1 mol NH₃)=8,25gNH₃

% Experimental

x ̅Masa=0,0988g
x ̅ = 31ml

%P/P=(((N*V)_NaOH-(N*V)_HCl )*PeqNH_3)/(g muestra)*100
%P/P=(((0,0788(eq-g)/L*0,05L)-(0,0801 (eq-g)/L*0,0310L))*17 g/(eq-g))/0,0988g*100
%P/P=25,06%


Tratamiento estadístico(aceptación o rechazo de resultados con el criterio Q)
#Eq=N*V=g/P_eq〖#EqNaOH〗_totales=〖#EqNaOH〗_reaccionan+〖#EqNaOH〗_(no reaccionan o en exceso)
〖(N*V)〗_NaOH 〖=#EqNH〗_3+#EqHCl
〖#EqNH〗_3=〖(N*V)〗_NaOH-〖(N*V)〗_HCl
gNH_3=(〖(N*V)〗_NaOH-〖(N*V)〗_HCl)/(PeqNH_3 )
%P/P=(((N*V)_NaOH-(N*V)_HCl)*PeqNH_3)/(g muestra)*100
1 Dato:
%P/P=(((0,0788 (eq-g)/L*0,05L)-(0,0801 (eq-g)/L*0,0313L))*17 g/(eq-g))/0,0994g*100
%P/P=24,5058%
2 Dato:
%P/P=(((0,0788 (eq-g)/L*0,05L)-(0,0801(eq-g)/L*0,0307L))*17 g/(eq-g))/0,0981g*100
%P/P=25,6634%
3 Dato:
%P/P=(((0,0788 (eq-g)/L*0,05L)-(0,0801 (eq-g)/L*0,0311L))*17 g/(eq-g))/0,0992g*100
%P/P=24,8298%
4 Dato:
%P/P=(((0,0788 (eq-g)/L*0,05L)-(0,0801(eq-g)/L*0,0309L))*17 g/(eq-g))/0,0985g*100
%P/P=25,2827%
Tratamiento estadístico (aceptación o rechazo de resultados con el criterio Q)
Ordenamiento de datos:
25,6634%
25,2827%
24,8298%24,5058%
x ̅ = 25,0704%

Valor mayor - x ̅ = 25,6634% - 25,0704%=0,5930%
x ̅- Valor menor = 25,0704% - 24,5058% = 0,5646%
Valor dudoso: 25,6634%
Valor cercano al dudoso: 25,2827%
R = Valor mayor –Valor menor
R = 25,6634% - 24,5058% = 1,1576%

Criterio de Q
Qexp= (∣Valor dudoso-Valor cercano al dudoso∣)/R
Qexp= (∣25,6634%-25,2827%∣)/(1,1576%)
Qexp=0,3289%

Qcrítica= 0,970

Qexp<...