Question

Consider the following unbalanced reaction. P4(s) + F2(g) PF3(g) How many grams of F2 are needed to produce 186 g of PF3 if the reaction has a 76.3% yield?

Answer #1

**The balanced chemical reaction is:**

**P4(s) + 6F2(g) --> 4PF3(g)**

**actual mass = 186 g**

**use:**

**% yield = actual mass *100 / theoretical
mass**

**76.3 = 186*100 / m**

**m = 243.8 g**

**Molar mass of PF3,**

**MM = 1*MM(P) + 3*MM(F)**

**= 1*30.97 + 3*19.0**

**= 87.97 g/mol**

**mass(PF3)= 243.8 g**

**number of mol of PF3,**

**n = mass of PF3/molar mass of PF3**

**=(243.8 g)/(87.97 g/mol)**

**= 2.771 mol**

**from balanced reaction,**

**moles of F2 required = (6/4)*moles of PF3
formed**

**= (6/4)*2.771 mole**

**= 4.16 moles**

**Molar mass of F2 = 38 g/mol**

**mass of F2,**

**m = number of mol * molar mass**

**= 4.16 mol * 38 g/mol**

**= 158 g**

**Answer: 158 g**

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