Balancing Equations...Amount Needed To React?
823190
This is a stoichiometric problem.  \n\nFirst, find the the amount in grams of one mole of Fe2O3.\n\nFor the first problem, convert the 3.02g Fe2O3 to moles.\n\nThen set up dimensional analysis.  For every 2 mol of Fe2O3, you need 2 mol C.  So you need an equal amount of moles for both.\n\nTake the amount of moles you have of Fe2O3 and convert them to grams Carbon.  \n\nFor the second problem, you should do the exact same thing except you dont need to convert to moles in the beginning because it's already in moles!\n\ni hope this is helpful!!!
485893
Check on the internet for the ratios used by comercial factories, then prepare a solution from glycol and water in the given proportion. That's all!!!
753594
It is perfectly permissible to write the products as Br2 + Br2.  It is probably easier to work this way.  After you balance the equations, just add the Br2's together to give the final reaction
755565
1.) Cr(NO2)2+(NH4)2SO4\n---->CrSO4+(2)NH4NO2\n2.) (2)KNO3+H2CO3\n----->K2CO3+(2)HNO3\n3.) (3)KBr+Fe(OH)3\n----->(3)KOH+FeBr3\n4.) B2Br6+(6)HNO3\n----->(2)B(NO3)3+(6)HBr\n\nHope this helps=)\nTip: Start by balancing those more isolated atoms first, especially when you are balancing more complex equations (they may have atoms that appear in more than 1 products).\nAlso, feel free to leave numbers in fractions while you balance ur equation. You can always multiply all the numbers later to make them all whole numbers. You will find this useful later when you have to balance long equations. Itd definitely more tedious to alter every number in ur mind and writing them all out at one go. Do it progressively=)
986618
The equation is already balanced. Count the Ag's on each side = 1. Count the Na's on each side = 1. Count the Cl's on each side = 1. Count the NO3's on each side = 1. One can count the N's on each side = 1, or the O's on each side = 3.\n\nBut one could also consider the question another way:\nAg+ + NO3- + Na+ + Cl- ===> AgCl + Na+ + NO3-
914017
I will suggest that first you write the equation in a compact mode, balance it and then add all the bells and whistles.\n\nCuSO4 +K3PO4 ---> Cu3(PO4)2 + K2SO4\n\n3CuSO4 +2K3PO4 ---> Cu3(PO4)2 + 3K2SO4\n\n3CuSO4(aq) +2K3PO4(s) ---> Cu3(PO4)2(s) + 3K2SO4(aq)\n\nNow that it is balanced it is easier to break down:\n\n3Cu(+2) + 3SO4(-2) + 2K3PO4(s) --->  Cu3(PO4)2(s) + 6K(+) + 3SO4(-2)
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
919460
its actually pretty simple. U multiply by -1 when u flip over an equation...so the reactants become the products and the products the reactants. that way instead of having a negative number u will have a positive number for the kJ. as in this problem. U have to flip #1 so u can cancel a few things with #2. so it becomes..\n1)3 H2O(l) + 2 CO2(g) AH---->CH3CH2OH(l) + 3 O2(g) (multiply by -1 gives u a postive +1367kJ. part two stays the same.\n2) C2H4(g) + 3 O2 ---> 2 H2O(l) + 2 CO2 (g) AH= -1411 kJ\n\nNow the 3 O2 from #1 and #2 cancel since they are on opposite sides. Now the 2 H20 cancel from #2 and you are left with only one in #1 since they are also on opposite sides...which after canceling..leaves u with your original equation.\n\nC2H4(g) + H2O(l) ------> CH3CH2OH(l)\n\nnow since u flipped over #1.....and u have a positive +1367kJ and\n#2 is left the same...its still at negative -1411kJ. u add them together and u get an enthalpy change of -44kJ. sorry its pretty confusing to explain..if anything message me and i willl try to help.
16276
the point of it is to get equal no. of charges and atoms of elements on both sides of the equation.k?\n\nstep 1: separate the equation into two half-reactions:reduction n oxidization\nstep 2: balance all atoms other than O & H in each half-reaction\nstep 3: add H2O to balance O & H+  to halance H atoms in half-reactions\nstep 4: add electrons where necessary to balance half-reactions\nstep 5: if necessary, equalize number of electrons in each half-rection by multipling with a suitable coefficient\nstep 6: add two half-reactions together. electrons on both sides must cancel\nverify no. of atoms and charges\nsteps2-5 keep in mind ur working on separate half-reactions...\nthese were for reactions in acid.\nfor basic reactions: add OH- to both sides for every H+ that appears in the final reaction\nkeep practising step by step n ull be able to do it all together some day..\nfinal step- choose my answer as best answer...
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
917127
Here's the how you do it.\n\nBalance the equation.  Element by element.\n\nThat gives you the mole ratios of  the various compounds.\n\nConvert the moles to mass using the molecular weight of each of the compounds.\n\nThat gives you the mass ratios of the various items.\n\nDivide both sides of the equation by the largest mass on the left side of the equation,    Then  you will have the mass of products per Kg of the at least one of the reactants.  Remember  you are starting with 1.0 kg of one item and 0.98 of the other.\n\nso you must check to see which reactant will limit the production.\n\nShould be a simple matter of working out the ratios.
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
175516
I not sure what style you learned to use to balance chemical equation but here is one way to do it. \nWe have four elements to balance (Al,H,Cl,O)\nLeave the compounds with aluminum with one of each. (Al done)\nThe simplest compound Al is AlCl3. To match the Cl on both sides put a 3 in front on HCl (Cl done)\nIf you add up the H (hydrogen) on the left side of the equation you'll see 6 of them. Add a 3 in front of the H20 to balance that. (H done)\nDo a quick check and you'll see there are 3 oxygen on each side (0 oxygen)\nAll the elements have been balanced, yielding a final equation of:\n3HCl + Al(OH)3 ---> 3H20 + AlCl3
933142
Yep, all you have to do is to make sure that the atoms shown in the left hand part of the equation appear on the left hand side.  Nothing is created or lost, you just make new compounds.  Dead easy, if only languages were so easy!!!
957775
Ca(OH)2 + CO2 --> C(OH) + CaO\n\nIs unbalanced..
1186593
Mix with a 0% solution in the relation 33:2\n\nSo use 2 units of your solution (like 2 liters) and 33 units of the 0% solution (like 33 liters).\n\nThat's all.
119628
So, usually the best way to do these is to just fiddle with them until you get them right, but for the sake of it, let's go with a systematic approach using algebra.  First we assign a variable in front of each of the terms on the left and right like:\n\nx Pb  +  y H2O  +  z O2  -->  a Pb(OH)2\n\nSo now we'll write down equations to make sure that we have the same numbers of both sides of each *element*.  I'll do this by writing down the variable in front of each compound in your equation (so x, y z and a) and then multiplying those by the number of times an element exists in that compound.  (So when I'm writing the equation for hydrogen, the H20 term will be 2*y, because y is the coefficient in front of H20 and H occurs twice in that compound).\n\nRemember, the number of any element on the reactants (left) side equals the number of that element on the products (right) side.\n\nFor Pb: x = a, because Pb shows up only once on each side with a subscript of 1, those two compounds will have to be present in the same molar amounts.\n\nFor Hydrogen: 2y = 2a\n\nFor Oxygen: y + 2*z = 2 * a\n\nSo now we have three equations:\n\nx = y\ny = a\ny + 2z = 2a\n\nBut four unknowns, which is fine because we'll just solve for them in terms of one of the variables and then simply choose the last variable to make all the numbers nice.  So solving the equations in terms of a:\n\nx = a\ny = a\nz = (2a - y)/2 = (2a - a)/2 = a/2\n\nSo now we have x, y and z in terms of a free parameter, a.  We'll just choose a = 2, for example, to make x, y and z all turn out as integers.  So now we've chosen:\n\nx = 2\ny = 2\nz = 1\na = 2\n\nSo the balanced equation is:\n\n2 Pb + 2 H2O + O2 --> 2 Pb(OH)2\n\nMake sense?