What volume of ammonia is formed when 10 dm^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The balanced chemical equation for the reaction is:$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$According to Avogadro's law,at constant temper

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(B) The balanced chemical equation for the reaction is:$N_{2(g)} + 3H_{2(g)} \longrightarrow 2NH_{3(g)}$According to Avogadro's law,at constant temperature and pressure,the volume ratio of gaseous reactants and products is equal to their stoichiometric mole ratio.From the equation,$1 \ volume$ of $N_2$ reacts with $3 \ volumes$ of $H_2$ to produce $2 \ volumes$ of $NH_3$.Given $10 \ dm^3$ of $N_2$ and $30 \ dm^3$ of $H_2$,the stoichiometric ratio is exactly $1:3$.Therefore,the volume of $NH_3$ formed is $2 	imes 10 \ dm^3 = 20 \ dm^3$.

What volume of ammonia is formed when $10 \ dm^3$ dinitrogen reacts with $30 \ dm^3$ dihydrogen at same temperature and pressure (in $dm^3$)?

Similar Questions

How many grams of $Cu(NO_3)_2$ are needed to obtain $1 \ g$ of copper? $($ Atomic mass: $Cu = 63.5, N = 14, O = 16$ $)$

The volumes of two $HCl$ solutions $A$ (Normality $= 0.5 \ N$) and $B$ (Normality $= 0.1 \ N$) to be mixed for preparing $2 \ L$ of $0.2 \ N \ HCl$ are:

The volume of $0.6 \, M$ $NaOH$ required to neutralize $30 \, cm^3$ of $0.4 \, M$ $HCl$ is ........ $cm^3$.

$24 \ g$ of $Mg$ metal reacts with water. How many moles of $H_2$ will be liberated? ............. $mol$

The weight of pure $NaOH$ required to prepare $250 \ cm^3$ of $0.1 \ N$ solution is ......... $g$. (in $g$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module