Question

What is the maximum amount of oxygen difluoride that can be produced when 0.55 mol of oxygen reacts with .90 mol of fluorine? Please show work

Answer 1

The balanced chemical reaction is:

2 F2 + O2 —> 2 OF2

2 mol of F2 reacts with 1 mol of O2

so,

0.55 mol of O2 requires 0.55*2 = 1.10 mol of F2

But we have only 0.90 mol of F2

so,

F2 is limiting reagent

we will use F2 in our further calculations

Form balanced chemical reaction

mol of OF2 formed = mol of F2 reacted

= 0.90 mol

Molar mass of OF2,

MM = 1*MM(O) + 2*MM(F)

= 1*16.0 + 2*19.0

= 54 g/mol

mass of OF2,

m = number of mol * molar mass

= 0.9 mol * 54 g/mol

= 48.6 g

Answer: 48.6 g