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.
1.) 6.0g Ti, 6.0g F2
2.) 2.6g Ti, 1.6g F2
3.) 0.231g Ti, 0.295g F2
1 Answer
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1.)
(6.0 g Ti) / (47.8670 g Ti/mol) = 0.1253 mol Ti
(6.0g F2) / (37.9968 g F2/mol) = 0.1579 mol F2
0.1579 mole of F2 would react completely with 0.1579 x (1/2) = 0.07895 mole of Ti, but there is more Ti present than that, so Ti is in excess and F2 is the limiting reactant.
(0.1579 mol F2) x (1 mol TiF4 / 2 mol F2) x (123.8606 g TiF4/mol) = 9.8 g TiF4
2.)
(2.6 g Ti) / (47.8670 g Ti/mol) = 0.05432 mol Ti
(1.6 g F2) / (37.9968 g F2/mol) = 0.04211 mol F2
0.04211 mole of F2 would react completely with 0.04211 x (1/2) = 0.021055 mole of Ti, but there is more Ti present than that, so Ti is in excess and F2 is the limiting reactant.
(0.04211 mol F2) x (1 mol TiF4 / 2 mol F2) x (123.8606 g TiF4/mol) = 2.6 g TiF4
3.)
(0.231 g Ti) / (47.8670 g Ti/mol) = 0.0048259 mol Ti
(0.295 g F2) / (37.9968 g F2/mol) = 0.0077638 mol F2
0.0077638 mole of F2 would react completely with 0.0077638 x (1/2) = 0.0038819 mole of Ti, but there is more Ti present than that, so Ti is in excess and F2 is the limiting reactant.
(0.0077638 mol F2) x (1 mol TiF4 / 2 mol F2) x (123.8606 g TiF4/mol) = 0.481 g TiF4
