m = 117.67 g of Ag(NO3) + Li(OH) reaction
m = 10.31 g of LiOH
the reaction:
Ag+ + OH- = AgOH(s)
mol of Ag+ = mol of AgNO3/MW of AgNO3 = (117.67)/(169.87) = 0.692706
mol of OH- = mol of LiOH /MW of LiOH = 10.31/23.95 = 0.43048
ratio is 1:1 so clearly, OH- is limiting the reaction;
solid forms: 0.43048 mol of OH- forms 0.43048 mol of AgOH
mass = mol*MW = 0.43048*124.87554 = 53.756 g of AgOH
b)
excess reactant will be: AgNO3,
mol of AgNO3 left = mol of AgNO3 initially- mol of AgNO3 reacted = 0.692706-0.43048 = 0.2622 mol of AgNO3 left
mass of AGNO3 = mol*MW = 0.2622*169.87 = 44.539 g left
c)
account for no. of moles:
Li+ will remain, since no reaction = mol of LiOH initially = 0.43048 mol of Li+
OH- will react with ag+ to form AgOH, it is limiting reactant so none is left = 0 mol of OH-
mol of Ag+ = mol of Ag+ left; previously calculated in B = 0.2622 mol of Ag+
mol of NO3- = initial amount will not change = 0.692706 mol of NO3-