Question

A gas contains a mixture of ${\mathrm{NH}}_3\left(\mathrm{g}\right)$and ${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$,both of which react withrole="math" localid="1650344995587" ${\mathrm{O}}_2\left(\mathrm{g}\right)$to form${\mathrm{NO}}_2\left(\mathrm{g}\right)$and${\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$.The gaseous mixture (with an initial mass of 61.00 g) is reacted with 10.00 moles${\mathrm{O}}_2$, and after the reaction is complete, 4.062 moles of${\mathrm{O}}_2$remains. Calculate the mass percent of${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$in the original gaseous mixture.

Short answer

The mass percent of ${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$in the original gaseousmixture is $87.33\%$.

Step-by-step solution

Definition

The mass percent of an atomin a compound is equal to the ratio of the atom's total mass to the complex'stotal mass multiplied by 100. As a result, it may be stated as follows:

Mass percent of an atom$=\left(\frac{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{atom}}{\mathrm{total}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{substance}}\right)100$

Determination of chemical reactions

The mass percent of ${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$in the initial gaseous mixture must be calculated.

A total of 10.00 mol of oxygen gas is reacted, leaving 4.062 mol when the reaction is complete.

As a result,

${\mathrm{n}}_{\mathrm{used}}\left({\mathrm{O}}_2\right)=5.938\mathrm{mol}$

Now, make a list of the chemical processes that take place:

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+7{\mathrm{O}}_2\left(\mathrm{g}\right)\to 4{\mathrm{NO}}_2\left(\mathrm{g}\right)+6{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+3{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NO}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Determination of initial mass of the mixture 

Let’s add the equations,

$4{\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+10{\mathrm{O}}_2\left(\mathrm{g}\right)\to 6{\mathrm{NO}}_2\left(\mathrm{g}\right)+8{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Whether ${\mathrm{n}}_{\mathrm{used}}\left({\mathrm{O}}_2\right)=5.938\mathrm{mol}$then,

$\mathrm{n}{\left({\mathrm{O}}_2\right)}_{\mathrm{used}}=\mathrm{n}\left({\mathrm{NH}}_3\right)+\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$5.938\mathrm{mol}=\mathrm{n}\left({\mathrm{NH}}_3\right)+\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

localid="1650345669292" $=\mathrm{x}+\mathrm{y}$

Thus the mass of the mixture initially was 61.00g,

${\mathrm{m}}_{\mathrm{mixture}}=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·\mathrm{m}\left({\mathrm{NO}}_2\right)+\mathrm{\omega }\left({\mathrm{N}}_2{\mathrm{H}}_4\right)·\mathrm{m}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$61.00\mathrm{g}=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·\mathrm{M}\left({\mathrm{NO}}_2\right)·\mathrm{x}+\left(1-\mathrm{\omega }\left({\mathrm{NO}}_2\right)\right)·\mathrm{y}·\mathrm{M}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

Determination of the value of x and y

The following ratios are derived from the chemical equation:

$\mathrm{n}\left({\mathrm{NH}}_3\right):\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)=4:1$

$\mathrm{n}\left({\mathrm{NH}}_3\right)=4\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$\mathrm{x}=4\mathrm{y}$
Now, compute the mass percent using all of the previous equations:

$\mathrm{n}{\left({\mathrm{O}}_2\right)}_{\mathrm{used}}=\mathrm{n}\left({\mathrm{NH}}_3\right)+\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$5.938\mathrm{mol}=\mathrm{n}\left({\mathrm{NH}}_3\right)+\mathrm{n}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$=\mathrm{x}+\mathrm{y}$

localid="1650345924371" $=4\mathrm{y}+\mathrm{y}$ $\because \mathrm{x}=4\mathrm{y}$

$=5\mathrm{y}$

$\mathrm{y}=1.19\mathrm{mol}$

$\mathrm{x}=4\times 1.19\mathrm{mol}$

$\mathrm{x}=4.75\mathrm{mol}$

Determination of the mass percent of N2H4.

Let’s add the molarities,

${\mathrm{m}}_{\mathrm{mixture}}=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·\mathrm{m}\left({\mathrm{NO}}_2\right)+\mathrm{\omega }\left({\mathrm{N}}_2{\mathrm{H}}_4\right)·\mathrm{m}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$61.00\mathrm{g}=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·\mathrm{M}\left({\mathrm{NO}}_2\right)·\mathrm{x}+\left(1-\mathrm{\omega }\left({\mathrm{NO}}_2\right)\right)·\mathrm{y}·\mathrm{M}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)$

$=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·46.01·4.75+\left(1-\mathrm{\omega }\left({\mathrm{NO}}_2\right)\right)·1.19·32.052$

$=\mathrm{\omega }\left({\mathrm{NO}}_2\right)·218.57+38.14-38.14\times \mathrm{\omega }\left({\mathrm{NO}}_2\right)$

$22.86=180.43\mathrm{\omega }\left({\mathrm{NO}}_2\right)$

$\mathrm{\omega }\left({\mathrm{NO}}_2\right)=0.1267$

$\mathrm{\omega }\left({\mathrm{N}}_2{\mathrm{H}}_4\right)=1-0.1267$

$=0.8733$

$=87.33\%$

Therefore the mass percentage of ${\mathrm{N}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)$is $87.33\%$.