Question 425b9

Jan 27, 2018

etaN_2=0.624mol; mN_2=17.5g
etaH_2O=1.25mol; mH_2O=22.5g#

Explanation:

1. Write and balance the equation
${N}_{2} {H}_{4} \left(l\right) + {O}_{2} \left(g\right) \to {N}_{2} \left(g\right) + 2 {H}_{2} O \left(g\right)$
2. Find the molar masses of the two reactants. Refer to the periodic table for the relative atomic masses of elements composing them.
${N}_{2} {H}_{4} = \frac{32.05 g}{m o l}$
${O}_{2} = \frac{32.00 g}{m o l}$
3. Given the masses of the reactants, find individual number of moles.
$\underline{\eta {N}_{2} {H}_{4}} :$
$= 20.0 \cancel{g {N}_{2} {H}_{4}} \times \frac{1 m o l {N}_{2} {H}_{4}}{32.06 \cancel{g {N}_{2} {H}_{4}}} = 0.624 m o l {N}_{2} {H}_{4}$
$\underline{\eta {O}_{2}} :$
$= 20.0 \cancel{g {O}_{2}} \times \frac{1 m o l {O}_{2}}{32.00 \cancel{g {O}_{2}}} = 0.625 m o l {O}_{2}$
4. Now, find the limiting reactant by multiplying each involved compound to its molar ratio; i.e.,
$\textcolor{red}{\underline{\eta {N}_{2} {H}_{4} \text{ available} = 0.624 m o l}} :$
$= 0.624 \cancel{m o l {N}_{2} {H}_{4}} \times \frac{1 m o l {O}_{2}}{1 \cancel{m o l {N}_{2} {H}_{4}}} = 0.624 m o l {O}_{2}$

This means that

$0.624 m o l {N}_{2} {H}_{4} \equiv 0.624 m o l {O}_{2}$, but

$\frac{\eta {O}_{2} \text{ available")/(0.625molO_2)>(etaO_2 " required}}{0.624 m o l {O}_{2}}$

$\textcolor{b l u e}{\underline{\eta {O}_{2} \text{ available} = 0.625 m o l}} :$
$= 0.625 \cancel{m o l {O}_{2}} \times \frac{1 m o l {N}_{2} {H}_{4}}{1 \cancel{m o l {O}_{2}}} = 0.625 m o l {N}_{2} {H}_{4}$
This means that

$0.625 m o l {O}_{2} \equiv 0.625 m o l {N}_{2} {H}_{4}$

$\frac{\eta {N}_{2} {H}_{4} \text{ available")/(0.624molN_2H_4)<(etaN_2H_4 " required}}{0.625 m o l {N}_{2} {H}_{4}}$

Therefore; $\textcolor{red}{\eta {N}_{2} {H}_{4} \text{ is the limiting reactant}}$

5.Now, find the products in grams using the limiting reactant and the mole ratios of the involved compounds as described in the balanced equation above; i.e.,

$\underline{{N}_{2} \text{ produced}}$
$= 0.624 \cancel{m o l {N}_{2} {H}_{4}} \times \frac{1 m o l {N}_{2}}{1 \cancel{m o l {N}_{2} {H}_{4}}} = 0.624 m o l {N}_{2} , t h e n$
$= 0.624 \cancel{m o l {N}_{2}} \times \frac{28.02 g {N}_{2}}{1 \cancel{m o l {N}_{2}}} = 17.5 g {N}_{2}$
$\underline{{H}_{2} O \text{ produced}}$
$= 0.624 \cancel{m o l {N}_{2} {H}_{4}} \times \frac{2 \cancel{m o l {H}_{2} O}}{1 \cancel{m o l {N}_{2} {H}_{4}}} = 1.25 m o l {H}_{2} O$
$= 1.25 m o l {H}_{2} O \times \frac{18 g {H}_{2} O}{1 \cancel{m o l {H}_{2} O}} = 22.5 g {H}_{2} O$