Best Answer: 2 LiCl = 2 Li + Cl2 mass Li = 56.8 mL x 0.534 g/mL=30.3 g moles Li = 30.3 g / 6.941 g/mol=4.37 the ratio between Li and LiCl is 2 : 2 ( or 1 : 1)
moles LiCl required = 4.37 mass LiCl = 4.37 mol x 42.394 g/mol=185.3 g
Cu + 2 AgNO3 = Cu/NO3)2 + 2 Ag the ratio between Cu and AgNO3 is 1 : 2 moles AgNO3 required = 4.2 x 2 = 8.4 : but we have only 6.3 moles of AgNO3 so AgNO3 is the limiting reactant moles Cu reacted = 6.3 / 2 = 3.15 moles Cu in excess = 4.2 - 3.15 =1.05
N2 + 3 H2 = 2 NH3 moles N2 = 42.5 g / 28.0134 g/mol=1.52 the ratio between N2 and H2 is 1 : 3 moles H2 required = 1.52 x 3 =4.56 actual moles H2 = 10.1 g / 2.016 g/mol= 5.00 so H2 is in excess and N2 is the limiting reactant moles NH3 = 1.52 x 2 = 3.04 mass NH3 = 3.04 x 17.0337 g/mol=51.8 g moles H2 in excess = 5.00 - 4.56 =0.44 mass H2 in excess = 0.44 mol x 2.016 g/mol=0.887 g
