Answer
177 g Ti, $1.41\times10^{3}\,g\,I_{2}$
Work Step by Step
Mass of Ti= $-1.55\times10^{3}\,kJ\times\frac{2\,mol\,Ti}{-839\,kJ}\times\frac{47.867\,g\,Ti}{1\,mol\,Ti}$
$=177\,g\,Ti$
Mass of $I_{2}$= $-1.55\times10^{3}\,kJ\times\frac{3\,mol\,I_{2}}{-839\,kJ}\times\frac{253.8\,g\,I_{2} }{1\,mol\,I_{2}}$
$=1.41\times 10^{3}\,g\,I_{2}$
