We must first identify the limiting reactant, and then we can calculate the theoretical yield.
We start with the balanced equation.
#color(white)(mmmmmmml)"Ti" +color(white)(l) "2F"_2 → color(white)(ll)"TiF"_4#
#"MM/g·mol"^"-1": 47.87color(white)(ll) 38.00color(white)(mll) 123.86#
(a) Identify the limiting reactant
We calculate the amount of #"TiF"_4# that can form from each reactant.
Calculate the moles of #"Ti"#.
#"Moles of Ti" = 4.8 color(red)(cancel(color(black)("g Ti"))) × "1 mol Ti"/(47.87 color(red)(cancel(color(black)("g Ti")))) = "0.100 mol Ti"#
Calculate moles of #"TiF"_4# formed from #"Ti"#
#0.100color(red)(cancel(color(black)("mol Ti"))) × "1 mol TiF"_4/(1 color(red)(cancel(color(black)("mol Ti")))) = "0.100 mol TiCl"_4#
Calculate the moles of #"F"2#
#"Moles of F"_2 = 3.2 color(red)(cancel(color(black)("g F"_2))) × "1 mol F"_2/(38.00 color(red)(cancel(color(black)("g F_2")))) = "0.0842 mol F"_2#
Calculate moles of #"TiF"_4# formed from #"F"_2#
#0.0842 color(red)(cancel(color(black)("mol F"_2))) × "1 mol TiF"_4/(2 color(red)(cancel(color(black)("mol F"_2)))) = "0.0421 mol TiF"_4#
The limiting reactant is #"F"_2#, because it produces fewer moles of #"TiF"_4#.
(b) Calculate the theoretical yield of #"TiF"_4#.
#"Theoretical yield" = 0.0421 color(red)(cancel(color(black)("mol TiF"_4))) × "123.86 g TiF"_4/(1 color(red)(cancel(color(black)("mol TiF"_4)))) = "5.2 g TiF"_4#
The theoretical yield of #"TiF"_4# is 5.2 g.
