The phosphorus in a $3\mathrm{.}014g$ sample of a plant food was converted to $P{O}_4^{3-}$ and precipitated as $A{g}_{3}P{O}_4$ by adding $40\mathrm{.}00mL$ of $0\mathrm{.}0672MAgN{O}_3.$ The excess $AgN{O}_3$ was back$-$titrated with $3\mathrm{.}15mL$ of $0\mathrm{.}0496MKSCN.$ Express the results of this analysis in terms of $\%P_2{O}_5.$

$P_2{O}_5+9H_{2}O2P{O}_4^{3-}+6H_3{O}^{+}$

$2P{O}_4^{3-}+6A{g}^{+}2A{g}_{3}P{O}_4(s)$

$A{g}^{+}+SC{N}^{+}$AgSCN$(s)$