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This method works great, but it seems to be limited for reactions with more than 4 components. It works when a reaction has 4 or fewer reactants/products, but the ABCD rule does not seem to extend to other E or F. Is there another method that does work on reactions with more than 4 components?
Ma'am according to my observation if value of A or B or C or D becomes a fraction then we have to make it whole number by multiplying the denominator with both the sides
I really like Byju's specially Kriti maaam .She teaches everything very deeply and clearly. Mujhe jyada English nhi aati so Byju's kya aaap Hinglish mein padha sakte hai please
Actually this is so long than the original method🥲 but I can use this to cross-check if necessary. And also is this method valid if there are more than 2 reactants and products ( like E and F is also there!?)
Ma'am I tried to use this trick for the following equation :- CO+O2 →CO2 However, the Trial And Error method is working for this equation, but the ABCD rule is not working here. It is resulting in a decimal value. Please help ma'am
aCO + bO2 --> cCO2 Trial and error is better for this one as it's a simple equation, but you can do a bit of algebra like this: C: a = c (1) O: a + 2b = 2c (2) Sub (1) in (2), c + 2b = 2c --> 2b = c cCO + (c/2)O2 --> cCO2 Then you can do some simple algebra: Divide through by c, and multiply through by 2 2CO + O2 --> 2CO2
@@cricup3835 Yes, you can do that. aNaHCO3 + bH2SO4 -> cNa2 + dSO4 + eCO2 + fH2O Na: a = 2c (1) -> c=a/2 H: a + 2b = 2f (2) C: a = e (3) O: 3a + 4b = 4d + 2e + f (4) S: b = d (5) Then sub for e and b in (4) using (3) and (5) 3a+4d = 4d + 2a + f -> f=a Now sub for f = (2) b = a/2 Then we have: a b=a/2 c=a/2 d=a/2 e=a f=a aNaHCO3 + a/2H2SO4 -> a/2Na2 + a/2SO4 + aCO2 + aH2O Then multiply through by 2 and divide through by a 2NaHCO3 + 1H2SO4 -> 1Na2 + 1SO4 + 2CO2 + 2H2O As the equations get more complex, linear algebra/matrix methods make it easier to solve a system of equations Also, you can alternatively recognise that SO4 is an ion and equate coefficients for SO4
Thank you, mam, for introducing that helpful shortcut during the lesson. The time-saving technique you demonstrated has proven invaluable in optimizing my workflow. I appreciate you taking the time to share your expertise and provide us with such a practical learning experience.
Mam, Thank you for helping me master such a unique and easy technique(far easier compared to our textbook techniques), it has truly been and will be helpful in the present and the upcoming times!
Divya mam,a awesome teacher.......❤ when she is teaching my whole mind is gonna concentrate on her.....😊..... I really proud to be a byju's student....... Vinay sir is so serious while teaching😅😊
This trick will NOT work for all the equations. Try balancing the equation of decomposition of lead nitrate. Also, at times, the equation might give a coefficient as negative or fraction.
