AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

In the formation reaction of ${NH}_{3}$ from $N_{2}$ and $H_{2}, 140 g$ $N_{2}$ and $40 {gH}_{2}$ were mixed. Select the option which is/are incorrect?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

If ${NH}_{3}$ formed is 85 gm then % yield will be 50%.

Maximum mass of ${NH}_{3}$ which can be formed is 180gm.

Some $N_{2}$ (g) will be left after the reaction.

If % yield of reaction is 80%, then $H_{2}$ consumed will be 32 gm.

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{lrl} N_{2} + 3 H_{2} ⟶ 2 {NH}_{3} (g) \\ 140 gm 40 gm \\ 5 mole 20 mole \\ LR \\ 0 5 mole 10 {mole}^{2} {NH}_{3} \end{array}$

(a) Mass of ${NH}_{3}$ produced = 170 gm

(b) Mass of $H_{2}$ used $= 30 \times \frac{80}{100} = 24 gm$

(d) Mass of ${NH}_{3}$ produce $= 170 \times \frac{50}{100} = 24 gm$

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

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

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!