2 Al + N2 -> 2 AlN Given mass of al= 2.7 g no of moles= given mass/ molar mass = 2.7/27=0.1 moles given mass of n2= 2.8 g no of moles= given mass / molar mass = 2.8/28= 0.1 moles no of moles of Al/ stoichiometric coefficient= 0.1/2 = 0.05 no of moles of N2/ stoichiometric coefficient= 0.1/1=0.1 0.05
No. Of moles (Al) = 2.7/27=0.1moles. No. Of moles (N2)=2.8/28=0.1moles. nAl/Al=nN2/N. 0.1/2=0.1/1 0.2=0.1 N2 - IT IS A LIMITING REAGANT . No.of mole = given mass / molar mass . 0.1×41/10=4.1 gm/mole [ANSWER].
@@ARYAN_120491.* Pehle dono ke moles nikalo 2. Fir pta kaise kre ki kon Limiting reagent hai tab uske liye 3. No.of moles ÷ stiochemetry coefficient 4. Then jo smaller ayega Al aur N2 me se wahi Limiting reagent hoga 5. Jo Limiting reagent hoga decide krega ki product kitna form hoga 6. Limiting reagent Al hoga 7.Aluminium ke 2 mole se 2 mole of AlN form hoga toh no. Of moles bhi same honge 8. Fir mass nikalne ke liye No. Of moles of Limiting reagent × Molecular mass of ALN which is 9. 0.1 × 41 = 4.1 gram * < Ans>
MAM APNE JO DIVIDE KARA HAI WO GALAT HAI 0.1/2=0.05 HOGA ,0.2 NHI HOGA . MAM YE APNE IDHAR CONFUSE KAR DIYA . APNE PAHLE BHI PADHYA HAI AUR YAHI GALTI TH I❤
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2 Al + N2 -> 2 AlN
Given mass of al= 2.7 g
no of moles= given mass/ molar mass
= 2.7/27=0.1 moles
given mass of n2= 2.8 g
no of moles= given mass / molar mass
= 2.8/28= 0.1 moles
no of moles of Al/ stoichiometric coefficient= 0.1/2 = 0.05
no of moles of N2/ stoichiometric coefficient= 0.1/1=0.1
0.05
If we take ALN molar mass 40
Then ans is 2 and if we take ALN 41..then ans is 2.05
Molar mass 40 wrong
answer is 4.1 gm
how ,,,plz explain
No. Of moles (Al) = 2.7/27=0.1moles.
No. Of moles (N2)=2.8/28=0.1moles.
nAl/Al=nN2/N.
0.1/2=0.1/1
0.2=0.1
N2 - IT IS A LIMITING REAGANT .
No.of mole = given mass / molar mass .
0.1×41/10=4.1 gm/mole [ANSWER].
how did u get 0.1/2=0.2 ?? it should be 0.05 ryt?? Al should be limiting regent not N2.
@mehakdeep777 if N2 is limiting reagent, then the answer should be 8.2g/mol
5.5 grams
answer 8.2 gram hai , stoichometric coefficient bhi lo N(ALN) ka formula use karte wakt
Homework answer is 4.1 gms
Same as your answer
kaise ?
*Ya bro you are right my answer is also the same as yours* 😊
@@ARYAN_120491.* Pehle dono ke moles nikalo
2. Fir pta kaise kre ki kon Limiting reagent hai tab uske liye
3. No.of moles ÷ stiochemetry coefficient
4. Then jo smaller ayega Al aur N2 me se wahi Limiting reagent hoga
5. Jo Limiting reagent hoga decide krega ki product kitna form hoga
6. Limiting reagent Al hoga
7.Aluminium ke 2 mole se 2 mole of AlN form hoga toh no. Of moles bhi same honge
8. Fir mass nikalne ke liye No. Of moles of Limiting reagent × Molecular mass of ALN which is
9. 0.1 × 41 = 4.1 gram * < Ans>
@@ashasingh8758thank you for explaining😊🤝
Answer will be 4.1 gm
why am i getting 8.2 gm?
4.1 g Homework answer
Hi. Yr ya ans kesa aaya.
Mam why you left pw
Homework answer 1.025 g
It help a lot as a jee aspirant .
8.2 answer hai
8.2 is right answer 4.1 aane ka sawal hi nhi paida hota
G mra 4.1 ans hai
Ur classes are so good mam ❤loved it 😊
Mass formed is 2.05 g. I hope my answer is not wrong😢😢😢
4.1 gram hoga
Your answer is incorrect
Mine is also this
Wtf
Correct ha
I got 4.1g as mass of AlN
AL is limiting reagent which tells that 4.1g of aln will form by mass
Pls explain bro ans kaise aaya
Answer is 4gm
Bhai 🤣 8.2 aaya hai yarr😂😂😂😂😂😂😂
@@SarthakMurmemera 4.1 g 😊
Thank for the lecture 🎉
Couls you Elaborate how you come to the expression of Stochiometry Cofficent And No of moles one in the video ?
Thank you for Nice Explanation 😊
Mam your last question is wrong acc. To law of conservation mass
Mass of product is equal to mass of reactants
Are thik to hai bhyi
9th nahi hai
AP ko dekh kar mind diverted ho gya❤
Answer is 4.1g ❤ mam this short revision lec are very usefull ❤😊
Topper,☢️😅🔥
Correct 👍😅
Answer is 2.05 grams.is it correct mam ❤❤❤❤❤❤❤❤❤
Yeahh
Galt h bhai
Home work answer is 4.1grams
Answer is 4.1 gm
Best example I have ever stood ❤❤❤❤😮😮😮😮
Thankyou mam for this amazing video i understand clearly after watching your video 📸
short and nice explaination mam thankyou
MAM APNE JO DIVIDE KARA HAI WO GALAT HAI 0.1/2=0.05 HOGA ,0.2 NHI HOGA .
MAM YE APNE IDHAR CONFUSE KAR DIYA .
APNE PAHLE BHI PADHYA HAI AUR YAHI GALTI TH I❤
Dude its o.1÷1/2 when 1 /2 will b converted into multiply it will b 0.1×2 /1 =0.2
But my ans 2.06g
nAlN=2.05
Ans 2.05 g
8.2 gram hoga mam
Are you mean "eight point two gram.
The answer for the last question is 4.1 grams
4.1g -ans for hw q
Thank you ma'am 💙
8.2g
answer is 4.1g
8.2 is right answer
Answer of last question is 4.1grams
Mam our channel is reached to 200k
Fantastic explanation mam🙏🙏
Mam answer is 4.1 g.. thanks for the session..can u please bring such videos on chemicsl bonding and structure
2g of AlN
Homework answer is 4.1 gm
🎉
8.2
The correct ans is 4.1 g
plz explain
Thank you mam❤❤❤❤
Mam Homework answer is 4.1 gram.
Ans - 4.1 g
Mam this is very helpful please complete all imp topics of physical che before neet
4.1gm are exactly answer we proved it
Only 4❤ mam tell me is it correct or not😊
Answer is 4.1 gram
8.2g is answer
Mam limiting reagent hai 0.05 and mass of AlN hai 4.1 plz tell me is it right or wrong❤
4.1
ryt
Mam aap kitni cute ho mera focus nhi ho pa rha video per isliye dobara dekh rha hu fir dobara bhi mera focus aapke cuteness per hi hai 😢
Mam I am studying this chapter from your lecture "wo 28 points bala" .. please tell me am I going in Correct direction
4.1 gram is the answer😅
Ans is 4.1g
Al is limiting
8.2gm
Can write these in school exam ? .
4.1 gram. thank you mam
I am first please pin me ❤❤
Mam please jee ke liye bhi padao Naa
Answer -- 8.2 grams ( exactly )
❤❤❤❤❤❤❤❤❤❤❤❤
Mine is also same , but comment mein max 4.1 bol rhe h , right kids h ??😢
Mam please percentage yield bhi samjha dijiye 🙏🙏
Ans is 4.1 grams ☺️👍♥️
4.1 gm
Thank you so much ma'am 😊
It's really helpful
Thanks 🎉🎉
Mam.. I've downloaded vision NEET app and registered also....
mam aap bohot acha padhate ho please again join pw
8.2gram❤
4.1❤
Superb lecture 👍👍👍👍👍
4.1 g is the correct answer ❤❤❤
Thanky you mam❤😊
Why we have to study the concept of linting reagent and excess reagent
Ma'am bio wali ma'am kaha gyi 🥺🥺🥺🥺🥺 unka teaching style awesome h 🙏🏻🙏🏻🙏🏻
2.75
4.1 gm answer
Ma'am agar limiting reagent ka ans same aajaye...
Toh jiski jyada stoichiometric value voh limiting reagent hoga na?
Nhi
@@krish10082
Then how will we decide? ...in that case
Thank you so much mam ❤
Very helpful video ❤
2.05
0.5125 answer
4.1 g ans
Home work ans 2.05.
Mam question practice plz❤
Mass= 4.1 gms
Hello ma'am
Very cute maam
Thankyou mam 🙎🏻🙎🏻
Hello mam
4.1gm answer 🎉🎉🎉🎉
❤😊
4.1 g
Homework and is 4.1 g of 2AlN are there
How is it 4.1 can u explain