Answer
$1.63$ $grams$ $PH_3$$BCl_3$
Work Step by Step
From 15.56a we found the concentration of $PH_3$ and $BCl_3$ to be $0.0432$ $M$.
The mole ratio for the stoichiometry of the whole equilibrium reaction is one to one. Therefore, the concentration of $PH_3$$BCl_3$ must be equal to $0.0432$ $M$.
The question asks what mass of $PH_3$$BCl_3$ will be required when contained in a $0.250$ $liter$ flask, to achieve the concentration (equilibrium) of $0.0432$ $M$.
Step 1:
$\frac{X mol PH_3BCl_3 }{0.250L }$ = $0.0432$ $\frac{mol}{L}$
Solve for $X$.
Step 2:
$Xmol$ $PH_3$$BCl_3$ = $0.0108$ $moles$ $PH_3$$BCl_3$
Step 3:
$0.0108$ $moles$ $PH_3$$BCl_3$ $\times \frac{151.33 grams PH_3BCl_3}{1 mol PH_3BCl_3}$ = $1.63$ $grams$ $PH_3$$BCl_3$
The answer comes to three significant digits.
