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The answers of first 2 equations are : 1. S + HNO3 -> H2SO4 + NO2 + H2O Solution • S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O solved equation -> Sulphur : a = c..(1), Hydrogen : b = 2c + 2e..(2), Nitrogen : b = d..(3), oxygen : 3b = 4c + 2d + e..(4) Let *b = 1 So, *d = 1 Now, from equation (2) and (4), we can write (b = 2c + 2e) and (3b = 4c + 2d + e) [1 = 2c + 2e] and [3(1) = 4c + 2 (1) + e] [1 = 2c + 2e] and [3 -2 = 4c + e] [1 = 2c + 2e] (5) and [1 = 4c + e] (6) ~ {2c + 2e = 4c + e} ~ {2c - 4c = -2e + e} ~ { -2c = -e } ~ { 2c = e } (substitution of e - class 10, chapter 3) Again from equation 6 1 = 4c + e { where e = 2c} 4c + 2c = 1 6c = 1 *C = 1/6 And for e = 2c e = 2 × 1/6 *e = 1/3 *a = c = 1/6 Now, after putting values, we will get -S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O 2. Cu + HNO3 -> Cu (NO3)2 + NO2 + H2O solution • Cu + 4HNO3 -> Cu (NO3)2 + 2NO2 + 2H2O - Copper : a = c..(1), Nitrogen : b = 2c + d..(2), Oxygen : 3b = 6c + 2d + e..(3), Hydrogen: b= 2e..(4) So, let *b= 1 *e = 1/2 (4) From equation (2) b = 2c + d, we get d = 1-2c (substitution of d - chapter 3 class 10) So in equation 3, we can write 3b = 6c + 2d + e (where d = 1-2c, e = 1/2, b= 1) 3(1) = 6c + 2(1-2c) + 1/2 3 = 6c - 4c + 5/2 ( after calculation) So, 2c + 5/2 = 3 *c = 1/4 [ this method is same as finding math's x😂] * a = c = 1/4 Now , d = 1-2c * d = 1/2 (after calculation) Hence, after adding the value of a,b,c,d,e We will get - Cu + 4HNO3 -> Cu (NO3)2 + 2NO2 + 2H2O #substitution is nothing but writing something simmilar which will be equal to 2nd value for solving 2nd value Like - 5 = b + c So we will find smth with variable b and equal to c - C = 2b So we can make this equation - in single variable - 5 = b + 2b, such that we can find the answer
S + 6 HNO3 ----------> H2SO4 + 6 NO2 + 2 H2O Cu + 4 HNO3 ------------> Cu ( NO3 ) 2 + 2 NO2 + 2 H2O Sir, a very big thanks to you 🙏 Watching after almost 4 years and this trick was amazingly perfect like magic 👌👌
@@dedgrim ABCD mein let A=1, abb ek equation solve kari for example B ki equation solve karni hai toh jo equation anti hai woh bana kar aagay bhadh jao nad C ke liye eqation bano agar uski bhi value nahi hain toh D ki bhi eqation bano and phir last mein ek equation dusri equation mein rakhne se value aa jaegi and sari value bhi nikal jaegi i know thoora sa problem hoga samjhne mein but yahan ese hi bata sakta houn
@@YuvrajSingh66666 A=1 let krnege to mushikil hora coz vo common bi nhi h. Charo me S,H,N& O Maine B ko 1 mana to Nitrogen me - b=d tha toh b ki value 1 and D ki value bhi 1 agyi. Than Oxygen me - 3b=4c+2d+e bach rha. To yaha pr mai puchra kaise niklega if i substitute b and d so I'll get - 3=4c+2+e ≈ 3-2=4c+e ≈ *1=4c+e* bas yaha se dikkt ari ab c and e ki value kaise nikale? Yhi same chez Hydrogen me bi hora - b= 2c+2e arha. Toh yaha Bi c and e ki value nikalni. ! How can i proceed plz tell?!!
I can't even describe how thankful I am to you,,, this was so much necessary for me. I was always confused with hit and trial. Now I have something logical and easy just because of you, thank you so much sir!!! 🥰
a = 1 b=6 c=1 d=6 e=2 we know that b = 2× 1 + e putting the value in 3b = 4c+2d +e also b = d there for we get e =2 putting e =2 in solution b =2c+2e we get b =6 and we know that b = d s d =6 and the equation formed is S + 6HNO3 => H2SO4 +6HNO2 + 2H2O this the correct solution for balancing it
Sir I am also a chemistry lecturer but seriously I didn't know about this trick but this trick helped me a lot. I used this trick to balance the equation in the classroom and the whole class room applauded me a lot just because of you...🤩
Sir pahle b=1 lege to d=1ho jayega aur phir hydrogen wale aur oxygen wale me elimination method laga denge isse e=1/3aayega aur c=1/6 and a=1/6 aur phir put karne ke baad multiply by 6 to equation balanced ho jaayegi
Elimination method means? I took b = 3 & e = 1,then i got a & c = 1/2 and d =3 . I took both e and b at a time and not with a value of 1 . Is this wrong method ? But i got right in both
I got a trick to balance any chemical equation with his and trial method and i solved all the equations of this video with that trick. The trick is very simple. You just have to make no. of all elements of products side even by multiplying 2 e.g a+b→ H2O(here no of O is odd) Then a+b→2H2O So that no. Of O and H become even. Now solve the equation of hit and trial method you will get the balanced equation
You are literally a genius sir, I watched almost 3 videos, and this was the 4th video I watched but none of the 3 were not explained well, I literally got this inside my head so goodly that I can't even forget this
Sir,This video Might have just made some EXTRA MARKS for my Papers that COULD HAVE BEEN LOST IF I WOULD HAVE NOT KNOWN SOME COMPLEX EQ BALANCING! Great Sir!
Hw Ist equation answers question S + HNO3 TO give H2SO4 + NO2 + H2O Answer S + 6HNO3 ------H2SO4 + 6NO2 + 2H2O Message: Thx for the homework it is easy i got b=d so i let c=1 and there is a equation in which 3 three variable are there b,d,e ,i also got b=d ,a=cand in the other equation there I got 2 variable after putting c=1 then in the equation where 3 variable exist I put b=d in RHS side where equation in which is put b=d is 3b=4+2d+e after putting b=d on rhs I got 3b=4+2b+e now just subract 2b from 3b then we have two equation Ist b=2+2end b=4+e multiply 2nd by 2 then eliminate e now you have your all the values
Bro jo carbon ya phosphorus ya sulphur ka oxidation hota h acids se voh jayadatar unke valence she'll ke electrons ke number se hi hota hai... Jaha tak meine class 10th observe Kiya hai...
Sir I just cannot express how happy I am after watching this video every other teacher have use the ABCD method but had not explained anything properly. You explain very well keep it up you earned a subscriber!
Given chemical equation, S+HNO3----->H2SO4+NO2+H2O Considering variable coefficients against each compound We get., aS+bHNO3------->cH2SO4+dNO2+eH2O On comparing number of atoms of each element on both the side one by one , Sulfur, a=c_____________(1) Nitrogen, b=d_____________(2) Oxygen, 3b=4c+2d+e___________(3) Let c=1, Then a=1 [from eq.1] Now, Putting b=2c+2e and c=1 and d=b in eq.2 We get, 3(2c+2e)=4×1+2(2c+2e)+e On further solving we get , e=2 Now putting c=1 and e=2 in eq.1 We get, b=6 As we know from eq.3 b=d Hence d=6 Now finally on putting all these values in our chemical equation we get, S+6HNO3-------->H2SO4+6NO2+2H2O This is the required balanced chemical equation. Am I right sir
I just started class 11th and chose pcm...my cousin who is 3 years elder to me just gave her hs examination and she's a very bright student...i always seek help in anything I face problem in especially in my studies. The first'thing she said me when I told her that I'm a little tensed about physics and chemistry, was that "Don't worry, just trust me and start learning from Physics Wallah, a yt channel". I started chemistry today and this is my first time on this channel and trust me guys, it's amazing!!!!!
S+6HNO3--- H2SO4+6NO2+2H2O is balanced equation.. It really works, great method... Solving with this method will always give correct answer..🙏🙏thakyou so much sir.....
The answers of first 2 equations are : 1. S + HNO3 -> H2SO4 + NO2 + H2O Solution • S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O solved equation -> Sulphur : a = c..(1), Hydrogen : b = 2c + 2e..(2), Nitrogen : b = d..(3), oxygen : 3b = 4c + 2d + e..(4) Let *b = 1 So, *d = 1 Now, from equation (2) and (4), we can write (b = 2c + 2e) and (3b = 4c + 2d + e) [1 = 2c + 2e] and [3(1) = 4c + 2 (1) + e] [1 = 2c + 2e] and [3 -2 = 4c + e] [1 = 2c + 2e] (5) and [1 = 4c + e] (6) ~ {2c + 2e = 4c + e} ~ {2c - 4c = -2e + e} ~ { -2c = -e } ~ { 2c = e } (substitution of e - class 10, chapter 3) Again from equation 6 1 = 4c + e { where e = 2c} 4c + 2c = 1 6c = 1 *C = 1/6 And for e = 2c e = 2 × 1/6 *e = 1/3 *a = c = 1/6 Now, after putting values, we will get -S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O 2. Cu + HNO3 -> Cu (NO3)2 + NO2 + H2O solution • Cu + 4HNO3 -> Cu (NO3)2 + 2NO2 + 2H2O - Copper : a = c..(1), Nitrogen : b = 2c + d..(2), Oxygen : 3b = 6c + 2d + e..(3), Hydrogen: b= 2e..(4) So, let *b= 1 *e = 1/2 (4) From equation (2) b = 2c + d, we get d = 1-2c (substitution of d - chapter 3 class 10) So in equation 3, we can write 3b = 6c + 2d + e (where d = 1-2c, e = 1/2, b= 1) 3(1) = 6c + 2(1-2c) + 1/2 3 = 6c - 4c + 5/2 ( after calculation) So, 2c + 5/2 = 3 *c = 1/4 [ this method is same as finding math's x😂] * a = c = 1/4 Now , d = 1-2c * d = 1/2 (after calculation) Hence, after adding the value of a,b,c,d,e We will get - Cu + 4HNO3 -> Cu (NO3)2 + 2NO2 + 2H2O #substitution is nothing but writing something simmilar which will be equal to 2nd value for solving 2nd value Like - 5 = b + c So we will find smth with variable b and equal to c - C = 2b So we can make this equation - in single variable - 5 = b + 2b, such that we can find the answer
Good evening sir , In the equation NH 3 + Cl 2→N 2 +NH 4 Cl We have done a NH 3 +b Cl 2 →c N 2 + d NH 4 Cl Then , Let d=1 → 2b - 1 b - ½ → 3a - 4 a - 4/3 → a - 2c + 1 4/3 - 2c 4/3*2 - c 4/6 - c 2/3 - c ̲ _̲_̲_̲_̲_̲_̲_̲_̲_̲ But sir in the video your answer coming is 1/6 But then also your answer is correct May you please answer sir ??
In this question S. a=c, H. b=2c+2e, N.b=d and O.3b=4c+2d+e.let c=1,then a=1 after that b=d=2+2e these value I put in the 3b=4c+2d+e then e=2 then final balanced equation is S+6HNO3=H2SO4+6NO2+2H2O
we teachers , most of time have learned all basic equations . ya but at tyms i use the first one , moreover in class 11 u have two more methods that are oxidation no method and ion e method , which though are lengthy but works very firm. as i have been teaching chem too since 3 years to class 11 n 12 so i mayuse those methods as well
Thank you sir you made this video,first we had a lot of problems in balancing but after watching this video of yours, all the problems themselves are running away from fear.👌👌👌👌👌☺️☺️☺️☺️☺️
Thankuuu so much sir. U re d best teacher I have ever seen.I'm in 10th class nd my boards exams r in 2021 nd now I m much confident to balance the equations.May u live long nd continue teaching us.
This is the first video that I am watching from physics wallah channel and now I understand why students appreciate this channel You are great sir 👌 I am so confused in this method but you solve my problem in just 15 min 😀 Thank you for this video 😊
answer of 1 hw A=c B= 2c+2e B= d 3b=4c+2d+e Let consider b=1 Now b=1 D=1 3(1)=4c+2(1)+e 3-2=4c+e 1=4c+e...........eq(1) B= 2c +2e 1= 2c+2e..........eq(2). ×2( because we have to make both 1 and 2 same) Now elemination method 4c+e=1 4c+4e=2 (-) (-) (-) ________ -3e=-1 E=1/3 Putting value of e in eq( 1) 4c+4(1/3)=1 C=1/6 A=c 1/6=c Now putting all value 1/6S+1HNO--- 1/6HSO+ 1NO 3 2 4 3 +1/3HO ×6 (2) Final ans 1S+6HNO3----1H2SO4+6N2O+2H20 I hope u all understand
