Question #d16b6

1 Answer

Answer:

#81.44%#

Explanation:

Mass of #Al(NO_3)_2#- #2.5kg#
Mass of #Mg(NO_3)_2#- #2.0kg#

-Conversion Factors-

Molar Masses:
Aluminum (Al)- #26.98g/(mol)#
Nitrogen(N)- #14.01g/(mol)#
Oxygen(O)- #16.00g/(mol)#
Magnesium(Mg)- #24.31g/(mol)#

#Al(NO_3)_2# molar mass
#=26.98g/(mol)+2(14.01g/(mol)+3*16.00g/(mol))#
#=151g/(mol)#

#Mg(NO_3)_2# molar mass
#=24.31g/(mol)+2(14.01g/(mol)+3*16.00g/(mol))#
#=148.33g/(mol)#

Conversion Equation
#Al(NO_3)_(2mass)*(Al(NO_3)_(2mols))/(Al(NO_3)_(2mass))*(Mg(NO_3)_(2mols))/(Al(NO_3)_(2mols))*(Mg(NO_3)_(2mass))/(Mg(NO_3)_(2mols))#
#=2.5kgAl(NO_3)_2*(1molAl(NO_3)_2)/(151gAl(NO_3)_2)*(1molMg(NO_3)_2)/(1molAl(NO_3)_2)*(148.33gMg(NO_3)_2)/(1molMg(NO_3)_2)#
#=2455.79gMg(NO_3)_2#

Percent Yield
#"%Yield"=("Actual")/("Theoretical")#
#=(2000gMg(NO_3)_2)/(2455.79gMg(NO_3)_2)=.8144=81.44%#