Question

Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous flourine: 12(s)+5F2(g) 2IFs(g) A 10.00 L flashia charged with 19.0 grams of l2 and 2.00 atm of F2 at 25°C. The flask is heated to 100°C until one of the reagents is completely consumed. What will be the total pressure (in atm) of the final products in the flask at 100°C? (Assume ideal gas behavior)

Answer 1

Molar mass of I₂=2xMolar mass of I=2x126.9 g/mol=253.8 g/mol

Number of moles of I₂=Given mass/Molar mass

=19.0 g/253.8 g/mol=0.075 mol

Pressure P of F₂=2.00 atm

Temperature T=25°C=25+273 K=298 K (0°C=273 K)

R=Gas constant=0.0821 Latm/molK

Volume V=10.00 L

n=number of moles of F₂

By PV=nRT

2.00 atm x 10.00 L=n x 0.0821 Latm/molK x 298 K

n=2.00 atm x 10.00 L/(0.0821 Latm/molK x 298 K)=0.82 mol

So we have 0.82 mol F₂ in the reaction mixture to begin with

As per the balanced chemical equation

1 mol I₂ reacts with 5 mol F₂

So 0.075 mol I₂ reacts with 5x0.075 mol=0.375 mol F₂

But we have 0.82 mol F₂ which is more than the required amount

So I₂ is the limiting reagent as it gets consumed completely

Now 1 mol I₂ produces 2 mol IF₂

0.075 mol I₂ produces 2x0.075 mol=0.150 mol IF₅

At the end of the reaction, the temperature T=100°C=100+273 K=373 K

Number of moles of IF₅=0.150 mol

Number of moles of F₂ that reacted with 0.075 mol I₂=0.375 mol

Remaining number of moles of F₂=0.82 mol-0.375 mol=0.445 mol

So total number of moles of gaseous reactants and products in the flask=number of moles of F₂ left+ number of moles of IF₅

=0.445 mol + 0.150 mol=0.595 mol

Using following values in PV=nRT

T=373 K

R=0.0821 Latm/molK

number of moles n=0.595 mol

V=10.00 L

Px10.00 L=0.595 mol x 0.821 Latm/molK x 373 K

P=0.595 mol x 0.0821 Latm/molK x 373 K/10.00 L=1.82 atm

Total pressure of the final products in the flask at 100°C=1.82 atm