Question

Ti(s)+2F2(g)→TiF4(s) compute the theoretical yield of the product (in grams) for each of the following initial amounts of reactants.

a) 7.0 g Ti, 7.0 g F2 Express your answer using two significant figures.

b) 2.4 g Ti, 1.8 g F2 Express your answer using two significant figures.

c) 0.230 g Ti, 0.284 g F2 Express the mass in grams to three significant figures.

Answer 1

Ti(s) + 2F2(g) = TiF4(s)

a)

moles Ti = 7.0 g / 47.88 g/mol = 0.146

moles F2 required = 2 x 0.146 =0.292

actual moles F2 = 7.0 g / 38 g/mol = 0.184 => limiting reactant

moles TiF4 = 0.184/2 = 0.092

Theoritical yield of the product = mass of TiF4 = 0.092 x 123.88 g/mol= 11.41 g

b)

moles Ti = 2.4 g / 47.88 g/mol = 0.050

moles F2 required = 2 x 0.050 =0.1

actual moles F2 = 1.8 g / 38 g/mol = 0.047 => limiting reactant

moles TiF4 = 0.047/2 = 0.0236

Theoritical yield of the product = mass of TiF4 = 0.0236 x 123.88 g/mol= 2.93 g

c)

moles Ti = 0.23 g / 47.88 g/mol = 0.0048

moles F2 required = 2 x 0.0048 =0.0096

actual moles F2 = 0.284 g / 38 g/mol = 0.00747 => limiting reactant

moles TiF4 = 0.00747/2 = 0.0037

Theoritical yield of the product = mass of TiF4 = 0.0037 x 123.88 g/mol= 0.46 g