Start by balancing the equation for the reaction between hydrogen $"H"_2$ and nitrogen $"N"_2$, which produces ammonia $"NH"_3$:
$color(darkgreen)(1) color(white)(.) "N"_2 (g) + color(darkgreen)(3) color(white)(.) "H"_2 to color(darkgreen)(2) color(white)(.) "NH"_3 (g)$
From the coefficients $(n("N"_2))/(n("H"_2))=(color(darkgreen)(1))/(color(darkgreen)(3))$
meaning that all of the nitrogen and hydrogen molecules available would have been converted to ammonia if supplied at a $1:3$ ratio; otherwise one of the reagents will be in excess.
The question states that, however,
$(n'("N"_2))/( n' ("H"_2))=3.5/5.0=7/10$;
$7/10 color(purple)(>) 1/3$
meaning that the species represented in the numerator- nitrogen $"N"_2$- is in excess; The number of moles hence the mass of ammonia produced shall be calculated from the quantity of the limiting reactant- $"H"_2$- available. Also, from coefficients in the balanced equation:
$(n("NH"_3))/(n("H"_2))=(color(darkgreen)(2))/(color(darkgreen)(3))$
$n("NH"_3)=(color(darkgreen)(2))/(color(darkgreen)(3))*n("H"_2)$
$color(white)(n("NH"_3))=(color(darkgreen)(2))/(color(darkgreen)(3))*3.50 color(white)(l) "mol"$
$color(white)(n("NH"_3))=2.33 color(white)(l) "mol"$
Take $17.031 color(white)(l) g*"mol"^(-1)$ as the molar mass $M$ of $"NH"_3$,
$m("NH"_3)=n("NH"_3)*M("NH"_3)$
$color(white)(m("NH"_3))=2.33 color(white)(l) color(red)(cancel(color(black)("mol"))) * 17.031 color(white)(l) g*color(red)(cancel(color(black)("mol"^(-1))))$
$color(white)(m("NH"_3))=39.7 color(white)(l) "g"$
