Question

How many liters of nh3 are needed to react completely with 30.0 l of no (at stp)? 4nh3(g) + 6no(g) → 5n2(g) + 6h2o(g)?

Answer 1

Answer:

20 l of $NH_3$

Explanation:

the reaction happens under standard conditions (stp) that mean it happens at a temperature of 298 k and 1 atm of pressure

Balanced equation

$4NH_3_(g)+ 6NO_{(g)} \longrightarrow 5N_2_{(g)}+ 6H_2O_{(g)}$

NO reacts completely means that it is the limit reagent.

This means that the reaction will occur until all the NO is consumed and becomes the products.

we find the moles of NO present in 30 l of compound with the ideal gas equation

$P= 1atm\\V= 30.0l\\r= 0.082 \frac{amt.l}{k.mol} (constant of ideal gases)\\T= 298k\\n=?\\\\PV=nrT\\n= \frac{PV}{rT} \\n=\frac{1atm.30l}{0.082\frac{mol.l}{j.mol}.298K } \\n= 1.22 mol NO$

$4NH_3_(g)+ 6NO_{(g)} \longrightarrow 5N_2_{(g)}+ 6H_2O_{(g)}$

By the stoichiometric coefficients, we know that for the reaction to happen we need 6 mol NO for every 4 mol $NH_3$

How many moles of $NH_3$ will be necessary to fully react 1.22 mol of NO

We apply a simple rule of three

$1.22 mol NO.\frac{4mol NH_3}{6 mol NO}=0.813 mol NH_3$

Now with the ideal gas equation we find the liters

$PV=nrT\\V= \frac{nrT}{P}\\V=\frac{0.082\frac{atm.l}{Kmol}. 0.813mol.298K}{1 atm} \\V= 20 l$

Answer 2

20.0 L
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