Answer
$33.2\text{ grams $KrCl_4$}$
Work Step by Step
Reaction:
$$Kr+2Cl_2\Rightarrow KrCl_4$$
Find the limiting reactant:
$PV=nRT$
$0.5*15.0=n(0.08206)(623)$
$n=0.147\text{ mol Kr}$
$1.5*15.0=n*0.08206*623$
$n=0.4401\text{ mol $Cl_2$}$
Kr is the limiting reactant.
$0.147\text{ mol Kr}*\frac{1\text{ mol $KrCl_4$}}{1\text{ mol Kr}}*\frac{225.6100\text{ grams $KrCl_4$}}{1\text{ mol $KrCl_4$}}=33.2\text{ grams $KrCl_4$}$
