can anyone balance this equation?
140559
Using the abc method… \n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nFe-->ax=2c------eq.1\nO-->ya+4b=12c+2d+e----eq.2\nH-->2b=2e----------eq.3\nS-->b=3c+d----------eq.4\n\nLet a=1\n\nFrom eq.1\nax=2c\nx/2=2/2c\nx/2=c-----eq.5\n\nForm eq.3\n2b=2e\n2/2b=2/2e\ne=b-----eq.6 \n\nFrom eq.2\nYa+4b=12c+2d+e\n1y+4b=12(x/2) +2d+b (Form eq.5&6)\nY+3b=6x+2d\n2d=y-6x+3b\nd=(y+3b-6x)/2------------eq.7\n\nFrom eq.4\nb=3c+d (but c=x/2)\nb=3(x/2) +d\nReplacing eq.4 by eq.7 we will get\nb=3x/2+(y+3b-6x) ½\nb= (3x+y+3b-6x) ½\n2[b=(y+3b-3x) ½]\n2b=y+3b-3x\n3x-y=b-------eq.8\n\nReplacing eq.6 by eq.8\ne=b\ne=3x-y\n\nReplacing eq.7 by eq.8\nd=(y+3b-6x)/2\nd=[y+3(3x-y)-6x]/2\nd=[y+9x-3y-6x]/2\nd= (3x-2y)/2\n\nFrom the above calculation we know that:\na=1, b=3x-y, c=x/2, d=(3x-2y)/2 & e=3x-y. \n\n[a]FexOy+[b]H2SO4 -->[c]Fe2(SO4)3+[d]SO2 +[e]H2O\nWhen replacing the values we’ll get:\n[1]FexOy+[3x-y]H2SO4 -->[x/2]Fe2(SO4)3+[(3x-2y)/2]SO2 +[3x-y]H2O
959197
I have factored it please refer to.\n\nFrom this solution can be found
979796
2H3PO4+6NaOH-------->6H2O+2Na3PO4\n\nThere has to be two moles of phosphoric acid and six moles of sodium hydroxide for it to balance out
851400
the first equ is balanced, b/c on the left side u have Pb-1, C-1 O-3 on the right side u have Pb-1 C-1 O-3 so since both side are equal it is balanced\n\nans for question 2 BeCl2 + 3O2------Be(ClO3)2\nBe-1                                  Be-1\nCl-2                                    Cl-2\nO-was 2 now 6                                     O-6
975769
Al(s) + MnO4- (aq) -----> MnO2 (s) + Al (OH)4- (aq)\n\nSplit the above into two half equations:\n\n1) Al(s)  --->  Al (OH)4- (aq)  and\n2) MnO4- (aq) -----> MnO2 (s)\n\nBalance 1) as follows:\n\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq) now balance for charge:\n1) 4OH^-  +  Al(s)  --->  Al (OH)4- (aq)  +  3e^-\n\nBalance 2) as follows:\n2) MnO4- (aq) ---> MnO2 (s)  +  2H2O (l) (turn xs O into H2O)\nNow balance with H^+:\n2) 4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\nnow balance charge:\n2) 3e^-  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)\n\nNow add 1) and 2) (electrons happen to be equal):\n4OH^-  +  Al(s)  +  4H^+  +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\nnow OH^-  +  H^+  ---> H2O:\n\n4H2O  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +  2H2O (l)  Al (OH)4- (aq)\n\nSubtract 2H2O from both sides:\n2H2O (l)  +  Al(s)    +  MnO4- (aq) ---> MnO2 (s)  +    Al (OH)4- (aq)\n\nI hope that helps. If you need more help, please email me or re post.
683249
Oh, come on, it doesn't get any easier than this!\n\nSnO2 + 2H2 --> Sn + 2H2O\n\nAbove is balanced. 1 Sn on both sides, 2 O on both sides and 4 H on both sides.\nYou need to work a little harder on these yourself.
23536
is your balace wrong?\nI only figure below balance:\nMnO2+4HCl = MnCl4+2H2O
123389
distribute the x to all terms inside the equation\n\n2x^2-26x
691296
use that at equilibrium : n.p = ni^2\nand from charge neutrality: \n\nn + NA = p + ND
980488
Hmm. Maybe all you need is the guideline to balance it?\n\nFor beginners, this is probably the best way. After some time, you can omit the unnecessary steps.\n\nFirst, list down the number of atoms of elements present on both sides of equation: reactants vs products. In this case, 1S,4O vs 1S,3O.\n\nSecond, find a common number of atom(s) of an element on either side. Knowing that there is 1S on both sides, the reacting O2 molecule will thus have a stoichiometric coefficient of 1/2. This will ensure 3O on either side.\n\nThird, multiply throughout with 2 since since fractions are not preferred.\n\nOf course, math enthusiasts usually get the hang of this faster by looking for common factor and this method is exceptionally effective (faster) when reactions become complex.\n\nBut the best way is still, of course, to practice XD
29264
Yuo can't solve this.Because it is a formula and not an equation..If u assign some value equal to this formula such as 0 u have the solution.
607601
I think you set the equation to zero then you have to factor it.  Go to the link below if you wanna know how to do it. OR look your your math book.
841292
When properly balanced, the coefficients read from left to right, 1,3,2,3. The balancing used by the other two people doesnt even balance correctly!!
724411
Sure...\n\nfirst we want to combine like terms on each side of the equals sign.  Think of the equals like the balance point on a scale... you can do anything on one side as long as you match it (keep it balanced) on the other!\n\nso we want to combine the variables and the constants...\n\n8a - 3a + 4 = -a -4 + 2\n5a + 4 = -a - 2\n\nnow we want all the variables on one side, and constants on the other:\n\n5a -(-a) + 4 - 4 = -a-(-a) - 2 - 4\n5a + a = -2 - 4\n6a = -6\n\nnow we just divide by 6 to get a alone:\n\n6a/6 = -6/6\na = -1\n\nHope this helps! :)
243814
although the equation is not complete and therefore the full answer can not be given. i believe this equation will need to be differentiated using the product of a product rule.
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
17656
Best technique is to look for where it is unbalanced and change it so it is balanced. Repeat until it is balanced.\n\neg. We need 2 hydrogen here on the left, so go for:\n\nMg + 2HCl -> MgCl + H2\n\nNow we need another 2 as we have 2 Cl on the left and one of the right.\n\nMg + 2HCl -> 2MgCl + H2\n\nNow we need to have 2 Mg on the left as we have 2 Mg on the right. So we get:\n\n2Mg + 2HCl -> 2MgCl + H2\n\nAll balanced so that is the answer.