Question

For the reaction 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)6.0 g Ti, 6.0 g F2 (express answer to two sig figs) b)2.3 g Ti, 1.4 g F2 (express answer to two sig figs) c)0.234 g Ti, 0.298 g F2 (express mass to three sig figs) Please show steps!

Answer 1

a)    Ti(s)+2F2(g)→TiF4(s)

Molar mass of Ti = 47.8670 g/mol

No of moles of Ti = 6/47.8670 = 0.125 mole

Molar mass of F2 = 38 g/mol

No of moles of F2 = 6/38 = 0.158 mole.

limiting reagent is F2

No of moles of TiF4 = 0.158/2 = 0.079 mole

molarmass of TiF4 = 123.861 g/mol

mass of TiF4produced = 0.079* 123.861 = 9.8 grams

b) Molar mass of Ti = 47.8670 g/mol

No of moles of Ti = 2.3/47.8670 = 0.048 mole

Molar mass of F2 = 38 g/mol

No of moles of F2 = 1.4/38 = 0.037 mole.

limiting reagent is F2

No of moles of TiF4 = 0.037/2 = 0.0185 mole

molarmass of TiF4 = 123.861 g/mol

mass of TiF4produced = 0.0185* 123.861 = 2.3 grams

c)
Molar mass of Ti = 47.8670 g/mol

No of moles of Ti = 0.234/47.8670 = 0.0049 mole

Molar mass of F2 = 38 g/mol

No of moles of F2 = 0.298/38 = 0.0078 mole.

limiting reagent is F2

No of moles of TiF4 = 0.0078/2 = 0.0039 mole

molarmass of TiF4 = 123.861 g/mol

mass of TiF4 produced = 0.0039* 123.861 = 0.48 grams = theoretical yield