Question

For the reaction shown, calulate the theoretical yield of the product in grams for each of the initial quantities of reactants. Ti(s) + 2F2(g) -->TiF4(s)               a) 1.0 g Ti; 1.0 g F2 b) 4.8 g Ti; 3.2 g F2 c) 0.388 g Ti; 0.341 g F2    **I'm having a hard time understanding the limiting reactant part the most.

Answer 1

Molar mass of Ti = 47.8 g/mol

Molar mass of TiF4 = 123.8 g/mol

Molar mass of F2 = 38 g/mol

a) 1.0 g Ti; 1.0 g F2

Ti(s) + 2F2(g) ------>TiF4(s)   

47.8 g 2 x 38 g

= 76 g

1 g ? = more than 1 g

But we have only 1 g of F2.

Hence, F2 is the limiting reagent.

Theoretical yield is calculated based on F2.

Ti(s) + 2F2(g) ------>TiF4(s)   

2 x 38 g 123.8 g

= 76 g

1 g ?

? = (1 g/ 76 g) x 123.8 g TiF4

= 1.63 g

Therefore,

Theoretical yield of TiF4 = 1.63 g

b) 4.8 g Ti; 3.2 g F2

Ti(s) + 2F2(g) ------>TiF4(s)   

47.8 g 2 x 38 g

= 76 g

4.8 g ? = more than 4.8 g

But we have only 3.2 g of F2.

Hence, F2 is the limiting reagent.

Theoretical yield is calculated based on F2.

Ti(s) + 2F2(g) ------>TiF4(s)   

2 x 38 g 123.8 g

= 76 g

4.8 g ?

? = (4.8 g/ 76 g) x 123.8 g TiF4

= 7.8 g

Therefore,

Theoretical yield of TiF4 = 7.8 g

c) 0.388 g Ti; 0.341 g F2

Ti(s) + 2F2(g) ------>TiF4(s)   

47.8 g 2 x 38 g

= 76 g

0.388 g ? = more than 0.388 g

But we have only 0.341 g of F2.

Hence, F2 is the limiting reagent.

Theoretical yield is calculated based on F2.

Ti(s) + 2F2(g) ------>TiF4(s)   

2 x 38 g 123.8 g

= 76 g

0.341 g ?

? = (0.341 g/ 76 g) x 123.8 g TiF4

= 0.555 g

Therefore,

Theoretical yield of TiF4 = 0.555 g