Diborane, ${\mathrm{B}}_2{\mathrm{H}}_6$ , can be prepared by the reaction;

$3{\mathrm{NaBH}}_4+4{\mathrm{BF}}_3⟶3{\mathrm{NaBF}}_4+2{\mathrm{B}}_2{\mathrm{H}}_6$

If this reaction has a 70 percent yield, how many moles of ${\mathrm{NaBH}}_4$ should be used with excess ${\mathrm{BF}}_3$ in order to obtain 0.200 mol of ${\mathrm{B}}_2{\mathrm{H}}_6$?