how do you balance this equation? CaCO3(s)+HCl(aq) --> CaC12 (aq)+CO2(g)+H2O?
824141
Start off with the original equation\n\nCaCO3(s) + HCl(aq) --> CaCl2(aq) + CO2(g) + H2O\n\n\nStart off by making the Ca equal on both sides, which is fortunately already done\n\nThen make the Cl equal on both sides by putting a 2 in front of HCL\nCaCO3(s) + 2HCl --> CaCl2(aq) + CO2(g) + H2O\n\nThen make the C's on both sides equal, which again, is already done\n\nNext make the O's equal on both sides, which yet again is already done.\n\nFinally you make the H's equal on both sides, which is done also. \n\nSo the answer to this whole question is to put a 2 in front of the HCl. (I'm guessing you meant to put CaCl2 instead of CaC12 in the products)\n\nThat was easy.
887756
Basic balancing of chemical reactions. You can always add CO2 O2 and H2O to either side. Start with the Carbon. balance the Oxygen last since it is contained in CO2 and H2O.\n\n1) 2 C6H6 + 15 O2 ==> 12 CO2 + 6 H2O\n2) C12H22011 + 12 O2 ==> 12 CO2 + 11 H20\n3) C3H6O + 4 O2 ==> 3 CO2 + 3 H2O\n4) C4H10O + 6 O2 ==> 4 CO2 + 5 H2O\n5) C4H10O + 6 O2 ==> 4CO2 + 5 H2O
874983
1. Molarity = concentration and it expressed as mol per dm^3\n    Change the mass to mole and the 851 ml to dm^3. Then find   \n    the ratio of the number of moles to volume in dm^3\n\n2. Soluble ionic compounds breaks up the water molecules in H^+ ions and OH^- ions. Sodium carbonate give Na^+ ions and CO3^2- ions......Other ions that are present are HCO3^- ions called hydrogen carbonate ions. I have found that a solution of sodium carbonate is alkaline: implying that there are more OH^- ions to H^+ ions, due to the formation of HCO^- ions.\n3. OH^- (aq) + H^+(aq) ----> H2O (l)\n4. H^+ ions (ie. hydrogen ions), OH^- ions (ie. hydroxyl ions),\n   Ca^2+ calcium ions and Cl^- chloride ions\n5. Hydrogen ions, hydroxyl ions. ammonium ions and sulphate ions.\n6. Actually the calcium metal reacts with water to give hydrogen gas and calcium hydroxide. At the same time, calcium reacts with nitric acid to give calcium nitrate, water and brown nitrogen dioxide gas.\nHence there is no simple ionic equation for your question\nYou may want to consider this one:\nCa^0  +  2H^+   + 2OH^-  -----> H2(g)  + Ca^2+  + 2OH^-\n7. Hydrogen ions, hydroxyl ions, iron (III) ions Fe^3+  and sulphate ion SO4^2-
746063
CO3^2- + 2H^+ ===> CO2 + H2O
986414
Start by balancing the equation:\n2Ca3(PO4)2 (s) + 6H2SO4 (aq)-----> 6CaSO4 (s) + 4H3PO4(aq)\n\nThen figure out the molar masses of each of the reactants:\nCa3(PO4)2 = 3(40.078) + 2(30.973762) + 8(15.9994) = 310.176724\n\nH2SO4 = 2(1.00794) + 32.065 + 4(15.9994) = 98.07848\n\nDivide each of your known masses of reactants by its respective molar mass to determine how many moles of each you have, and divide this number by the coefficient (the number in front) of the reactant to determine how many "parts" of each you have:\n(1000/310.176724)/2 = 1.612 parts 2Ca3(PO4)2\n(980/98.07848)/6 = 1.665 parts 6H2SO4\n\nTake the smaller number (since once you run out of one reactant, the reaction is over) and figure out how many moles of calcium sulfate and phosphoric acid you'll produce:\n(1.612)(6)(40.078 + 32.065 + 4(15.9994)) = 1316.7518832g CaSO4\n\n(1.612)(4)(3(1.00794) + 30.973762 + 4(15.9994)) = 631.872933536g H3PO4\n\nSince the total mass of reactants and products needs to be the same, you can check your (er, my) work by totalling up both sides:\n(1.612)(2)(310.176724) + (1.612)(6)(98.07848) = 1948.625g\n1316.7518832 + 631.872933536 = 1948.625g
749125
The unbalanced equation is:\nCH3-OH + O2 (+ heat) ---> H2O + CO2\n\nThe balanced equation would be:\n2CH3-OH + 3O2 (+ heat)  ---> 4H2O + 2CO2
946866
11/27:Please post the answer so I can find the error in my work or determine if the answer key is wrong.  Lets see if we can get to the bottom of this one!\n\n\nFirst find the masses of each acid:\n\nHCl:  48.8 ml x 1.19 g/ml x 0.38 = 22.1 g\n\nHNO3:  17.5ml x 1.42 g/ml x 0.70 = 17.4 g\n\nSecond, convert grams to moles:\n\nMass HCl = 35.45 + 1.01 = 36.46 = 36.5 g/mol\nMass HNO3 = 16.0 x3 + 14.0 + 1.0 = 63.0 g/mol\n\nmoles HCl = 22.1 g/36.5 g/mol = 0.605 mol HCl\nmoles HNO3=17.4g/63.0 g/mol = 0.276 mol HNO3\n\nSince you're diluting to one liter, these are also the molarities.\n\n[HCl] = 0.605 mol/L ; [HNO3] = 0.276 mol/L\n\nEach of these acids are strong acids and can be assumed to dissociate completely.  If they were weak acids we would have to account for the acid dissociation equilibria here.  \n\nEach of these acids contribute only 1 equivalent of hydrogen ion.  Thus [H+] is the sum of the two concentrations.\n\n[H+] = [HCl] + [HNO3] = 0.605 + 0.276 = 0.881 mol/L = 0.88 M\n\nReport two sig figs since your mass % vals only have two sig figs.\n\nThe equilibrium equation for water, H+ and OH- is:\n\nH2O <--->  H+  + OH-   K = 1.00 x 10^-14\n\nH2O is a pure liquid and by definition has an activity of unity, so the equilibrium expression is:\n\n[H+][OH-] = 1.00 x 10^-14\n\nNow we can solve for [OH-] :\n\n0.881 X [OH-] = 1.00 x 10^-14\n\n[OH-] = 1.14 x 10^-14 mol/l = 1.1 x 10^-14 M\n\nAgain, two sig figs in final answer.\n\nFinally, pH = - log [H+]\n\nso pH = - log 0.881 = 0.055
839194
The balanced equation is\n\nH3PO4 + 3KOH ----> K3PO4 + 3H2O\n\nApart from using the rule to balance this, you can also view this in this way, H3PO4, when dissolved in solution, gives 3H+ and 1PO43-, while KOH when dissolved in solution gives 1K+ and 1OH-. OH- and H+ always neutralize each other to give water. Therefore, you need 3 OH- to neutralize the 3H+ from the acid. Work out the balance, you get the above equation.\n\nGenerally, number of atom and particles can be found by first finding the number of moles. Number of moles by definition is the mass of the substance divided by the molecular weight. For example, total number of mole in 12 g of C is \n12 / 12 = 1 mole\n\nWithin each mole, there are always a finite number of UNIT, which can be in the form of atom, molecule, or ion. This special number is called the Avogadro constant, and has a value of 6.02e23.\n\nFor example, 1 mole of C has 6.02e23 carbon atom.\n\nHowever, for 1 mole of H2, there are 6.02e23 H2 MOLECULE!!! because every two H atoms are bond together to form a more stable molecule. To find the total number of atom, you know each H2 molecule has 2 atom, so the total number of atom in this case is 2*6.02e23 or 12.04e23 atom.\n\nUsing the relationship\n\n# mole = mass/ molecular weight\n\nYou get mass = # mole * molecular weight
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
797533
Providing that the H3C6H5O7 is citric acid, that balanced equation is:\n\nH3C6H5O7 (s) + 3 NaOH (aq) --> Na3C6H5O7 (aq) + 3 H2O (l)\n\nThe net ionic equation is:\n\n3 H+ (aq) + 3 (OH-) (aq) --> 3 H2O (l)
957905
the balanced reaction\nC12H22O11 + 12 O2 ---> 12 CO2 + 11 H2O\nso the right answer are:\nc. \nand\ne.
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
764101
In all reactions one uses that 2 OH -> H2O + O, the split of oxygen goes to create the ion. Finally balance the equation:\n \nSO3 + 2 NaOH -> Na2SO4 + H2O\n\nCO2 + 2 KOH -> K2CO3 + H2O \n\nSO2 + Ba(OH)2 -> BaSO3 + H2O
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4