what is the equation for sulphuric acid and magnesium hydroxide?
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
103867
NaClO + 2FeSO4 + H2O -> NaCl + 2 Fe(3+) + 2 (SO4)(2-) + 2 OH(1-)
311033
H2SO4 - H20 = SO3\n\nAnyhydride is the removal of the atoms of water, therefore leaving Sulphur Trioxide.
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
691254
Magnesium sulfate = Epsom salt\n\nMagnesium hydroxide, Mg(OH)2, otherwise known as milk of magnesia\n\nIron(III) oxide — also known as ferric oxide, Hematite, red iron oxide, synthetic maghemite, colcothar, or simply rust
727675
Magnesium reacts with water to produce hydrogen.  It's salts are water soluble.
705101
Ferrous Hydrogen Sulfate is also called Iron (II) Hydrogen Sulfate.\n\nFe(HSO4)2  although FeH2(SO4)2 would also be correct.\n\nThe Hydrogen Sulfate anion is a Sulfuric acid with a single proton (Hydrogen) removed.  It has the formula:\n\n(HSO4)-1\n\nIt takes two of these to balance the charges on the Ferrous (+2)cation.
787481
NaHCO2 + H2SO4 --> Na2SO4 + CO2(g) + H2O(L)\n\nSodium bicarbonate + sulphuric acid equals sodium sulfate + carbon dioxide + water
727004
(NH4)2Cr2O7 -> Cr2O3 + 4*(H2O) + N2\n\n(NH4)2SO4 + 2*(NaOH) -> 2*(NH3) + 2*(H2O) + Na2SO4\n\nFe2(SO4)3 + 6*(NH3) + 6*(H2O) -> 2*(Fe(OH)3) + 3*((NH4)2SO4)
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.
899322
First find how many moles of each reactant you have\n\nmoles BaI2 = M*V= 1.35*0.055\nmoles MgSO4 = M*V= 1.2*0.0476\n\nThe reaction is \nBaI2 + MgSO4 ->BaSO4 + MgI2\n\nthe stoichiometry is 1:1, so you should have equal moles of the two reactants. In your case however one of them is in excess. So you will have to calculate the yield based on the limiting reagent.\n\nSince the stoichiometry is also 1:1 for BaSO4 and the limiting reagent you should have moles of BaSO4 = moles of the limiting reactant. But moles=mass/MW => mass= moles*MW\nYou have calculated the moles, find the MW of BaSO4 and you'll get the theoretical mass of BaSO4 that would come from 100% reaction.\nThe percent yield is (actual mass/theoretical mass)*100%
122744
Mg + 2 CH3COOH = Mg(CH3COO)2 + H2
803450
MgCl2(aq) + 2NaOH(aq)  --> Mg(OH)2(s) + 2NaCl(aq)\n\nNet ionic equation:\n\nMg2+(aq) + 2OH-(aq) --> Mg(OH)2(s)
277253
MgCl + H2O = MgO + H2Cl
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
326922
Another easy reaction:\n\nMg(s) + HNO3 ===========> Mg(NO3)2 + H2 (g)\n\nMagnesium reacts rapidly with nitric acid to make magnesium nitrate and hydrogen gas
424592
H2SO4 + 2NaOH ----> 2H2O + Na2SO4\n\nYOU MUST SEPARATE AND CANCEL THE COMMON IONS (AQUEOUS ONES)