91Ó°ÊÓ

We need to find the balanced equation.

We know that

The balanced equation is that in which the total mass of reactants is equal to the total mass of products.

Therefore, The balanced equation is,

$2{\mathrm{NH}}_3+5{\mathrm{F}}_2→{\mathrm{N}}_2{\mathrm{F}}_4+6\mathrm{HF}$

We need to find the no. of moles of each reactant are needed to produce $4.00$moleof $\mathrm{HF}$

From part (a)

We know that

The $6\mathrm{mole}$of $\mathrm{HF}$is produced from $2$moleof ${\mathrm{NH}}_3$and $5$moleof ${\mathrm{F}}_2$

Therefore, $4$moleof $\mathrm{HF}$is produced from localid="1653484659519" $\frac{2\mathrm{mole}}{6\cancel{\mathrm{mole}}}×4\cancel{\mathrm{mole}}=1.33\mathrm{mole}$of ${\mathrm{NH}}_3$and localid="1653484677099" $\frac{5\mathrm{mole}}{6\cancel{\mathrm{mole}}}×4\cancel{\mathrm{mole}}=3.33\mathrm{mole}$of ${\mathrm{F}}_2$

We need to find the mass of ${\mathrm{F}}_2$needed to react with $25.5$gof ${\mathrm{NH}}_3$

From part (a)

We know that

The $34$g of ${\mathrm{NH}}_3$reacts with $190$gof ${\mathrm{F}}_2$

Therefore, $25.5$gof ${\mathrm{NH}}_3$reacts with localid="1653484700702" $\frac{190\cancel{\mathrm{g}}}{34\cancel{\mathrm{g}}}×25.5\mathrm{g}=142\mathrm{g}$

We need to find the mass of${\mathrm{N}}_2{\mathrm{F}}_4$produced when $3.40$gof ${\mathrm{NH}}_3$reacts

From part (a)

We know that

The $34$gof ${\mathrm{NH}}_3$produces $104$gof ${\mathrm{N}}_2{\mathrm{F}}_4$

Therefore, $3.40$gof ${\mathrm{NH}}_3$produces localid="1653484719664" $\frac{104\mathrm{g}}{34\cancel{\mathrm{g}}}×3.4\cancel{\mathrm{g}}=10.4\mathrm{g}$of ${\mathrm{N}}_2{\mathrm{F}}_4$