Answer:
2.4 grams of ClF3
Explanation:
First let us determine the moles of Cl2 and F2,
Cl2 = ( ( 337 )( 2.05 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),
Cl2 = ( 690 / 24.436 ) * ( 1 / 780 ),
Cl2 = ( About ) 0.036 moles of Cl2
_________________________________________________
F2 = ( ( 729 )( 2 L ) / ( 0.082 )( 298 K ) ) * ( 1 atm / 780 ),
F2 = ( 1458 / 24.436 ) * ( 1 / 780 )
F2 = ( About ) 0.078 moles of F2
Now let us identify the limiting reactant, considering the ratio between ClF3 and Cl2 / F2. In this case F2 is the limiting reactant, as it forms a smaller molar ratio;
The theoretic yield is thus performed with the limiting reactant F2,
0.078 * ( 2 / 3 ) * ( 92.45 / 2 ) = ( About ) 2.4 grams of ClF3
