According to the equation,N2O3(g)+6H2(g)⟶2NH3(g)+3H2O(g)how many moles of NH3(g) could be formed from the reaction of 0.22 mol of N2O3(g) with 0.87 mol of H2(g)?

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

According to the equation,

$N_{2} O_{3} (g) + 6 H_{2} (g) ⟶ 2 {NH}_{3} (g) + 3 H_{2} O (g)$

how many moles of ${NH}_{3} (g)$ could be formed from the reaction of 0.22 mol of $N_{2} O_{3} (g)$ with 0.87 mol of $H_{2} (g)$ ?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

0.73 mol

0.29 mol

0.44 mol

1.1 mol

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In the reaction

$N_{2} O_{3} + 6 H_{2} \to 2 {NH}_{3} + 3 H_{2} O$

Here, number of moles formed of ammonia,

$\begin{aligned} \frac{0.87}{6} \times 2 \\ = 0.29 mol \end{aligned}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE