Question

For the reaction $N{H_3}\, + \,{O_{2\,}} \to \,NO + \,{H_2}O$ how many grams of ${O_{2\,}}$ are needed to completely react with $85.15g\,of\,N{H_3}$ ?

Answer 1

Firstly balance the chemical reaction given above, then see how many moles of ammonia reacts with oxygen. In these types of reactions, generally we have $1:1$ between the two reactants and thus after applying the mole concept the same amount of one reactant reacts with the other one. Balance and see the molar ratio here. Mole ratio may be other than just $1:1$ .

Complete step by step answer:
Firstly we have to balance the given chemical reaction. To balance it, we multiply the oxygen on the left hand side by $5$ it will make a total of $10\,oxygen$ on the left hand side $5{O_2}$ . Then we will balance oxygen atoms of the right hand side by multiplying ${H_2}O$ by $6$ . This will make oxygen somewhere nearer to oxygen of left hand side i.e. $5{O_2}$ but now we have to balance other atoms, we get $12\,hydrogen$ on right hand side, balance hydrogen on left hand side by multiplying ammonia with $4$ this will balance hydrogen, at last balance remaining atoms. Here only nitrogen is left to balance which on multiplying $NO$ by $4$ balances the whole equation.
We get, $4N{H_3}\, + \,5{O_{2\,}} \to \,4NO + \,6{H_2}O$
Now let’s proceed to the main part of our question that is to find the amount of ${O_{2\,}}$ to react with $85.15g\,of\,N{H_3}$ .
Here the molar ratio of $4:5$ is in between $N{H_3}\,$ and ${O_{2\,}}$ it means $4\,mole\,of\,N{H_3}$ are needed to react with $5\,mole\,of\,{H_2}O$ . We have $85.15g\,of\,N{H_3}$ thus it is equivalent to $5\,mole$ thus we need $\dfrac{5}{4}mole\,of\,oxygen$ according to the equation stoichiometry given below.
$N{H_3}\, + \,\dfrac{5}{4}{O_{2\,}} \to \,NO + \,\dfrac{6}{4}{H_2}O$
Calculating the exact moles of ammonia present here by the formula of number of moles, number of moles are the ratio of given mass to molar mass.
$moles\,of\,N{H_3} = \dfrac{{given\,mass\,of\,N{H_3}}}{{molar\,mass\,of\,N{H_3}\,}}$
$moles\,of\,N{H_3} = \dfrac{{85.15g}}{{17.03\,g\,mo{l^{ - 1}}\,}}$
$moles\,of\,N{H_3} = 5\,moles$
So, now calculating amount of oxygen by multiplying $\dfrac{5}{4} \times \,5\,moles\, \times molar\,mass\,of\,oxygen$
Molar mass of oxygen is $32$ , how it will be this for a single oxygen it will be $16$ and for ${O_{2\,}}$ it will $32$
$\dfrac{5}{4} \times \,5\, \times \,32\,g$
$200\,g$

We get $200g$ of oxygen needed for the completion of reaction.

Note: Balance chemical equation first and check for it whether it is balanced or not. We need to see the molar ratio between the reactants which gives us an idea about how much part we need to take for reacting oxygen with ammonia. Here we multiply molar mass of oxygen with $\dfrac{5}{4}$ according to the stoichiometry and by $5\,$ due to moles of ammonia.