How do I calculate the moles of Mg reacted?
609634
This question cant be answered without knowing the molarity of the HCl. I seriously doubt you were using 12M as another poster suggested. \n\nAlso. if all the Mg reacted the moles of Mg reacted would simply be 0.15/24.3 = 0.006. (Did you really use 0.15g? Thats a small amount) \n\nIf you didnt actually preform the rxn and/or dont know if all the Mg reacted you need to use the moles of Mg calculated above, and compare it to the moles of HCl used. M = moles / liters. So plug in the molarity of HCl used, and 0.050 L (converted 50mL) and solve for moles of HCl. Next use the mole ratio to determine the limiting reagent. If you dont know how to do that, your screwed.
778417
Find the moles of AgNO3:\n\nn = C*V, n = 0.14*25x10^(-3) mol\n\nAccording to the chem. equation:\n\n2AgNO3 + Na2S  --> 2NaNO3 + Ag2S\n\n2 moles of AgNO3 reacts fully with 1 mole of Na2S, so\n0.14*25x10^(-3) moles of AgNO3 reacts with 0.07*25x10^(-3) moles of Na2S\n\nNow you can find the volume of the Na2S solution:\n\nn = C*V  =>  V = n/C = 0.07*25x10^(-3)/0.105 = 0.05/3 L or\n\nV = 16.7 mL (approx)
958410
1.\nMn(s) + SnCl2(aq) --> MnCl2(aq) + Sn(s)\nCr(s) + 2HBr(aq) --> CrBr2(aq) + H2(g)\nCo(s) + CaBr2(aq) --> N.R\n2Al(s) + 3Ni(NO3)2(aq) --> 2Al(NO3)3(aq) + 3Ni(s)\n\n2.\nMg + 2AgNO3 --> Mg(NO3)2 + 2Ag\nmole of Ag = 3.580 / 108 = 0.033 mole\nmole of AgNO3 = 0.033 mole = 33 mmole\nmolarity of AgNO3 = 33 / 50 = 0.66 M\n\n3.\na.soluble\nb.soluble\nc.soluble\nd.insoluble\ne.soluble\nf.soluble\ng.insoluble in water, soluble in acid\nh.soluble\n\n4.\n3NaOH(aq) + Fe(ClO4)3(aq) --> Fe(OH)3(s) + 3NaClO4(aq)\n3Na+ + 3OH- + Fe3+ + 3ClO4- --> Fe(OH)3 + 3Na+ + 3ClO4-\nFe3+ + 3OH- --> Fe(OH)3\n(NH4)2SO4(aq) + Pb(NO3)2(aq) --> PbSO4(s) + 2NH4NO3(aq)\n2NH4+ + SO42- + Pb2+ + 2NO3- --> PbSO4 + 2NH4+ + 2NO3-\nPb2+ + SO42- --> PbSO4\n\nHope it helps\npeace\nvixklen
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
955628
For the reaction:\n\nMgO + 2 HCl  -->  MgCl2 + H2O\n\nThe temperature should go up, not down.  Only the initial and final temperature are needed.\n\nTypical values for students are between 110 an 150 kJ/mol MgO.  This is typical, not the best that can be done.
697545
1- find heat quantity\n Q=C*t\nQ = heat quantity of the reaction\nC= heat capacipy of calorimeter\nt=temperature increasing\n\n2- find heat of reaction \n H=(Q*1000)/(N*V)\nN = normality of acid and base\nV= volume of acid and base
206920
lime = CaCO3; CaC03 + 2HCl --> CaCl2 + CO2 + H2O. You will get 1 mole of CO2 (44g) for every mole of lime reacted.
924159
It depends on what you have, if you know the mole of the molecule, you can determine the mass by converting molar mass (g/mol) into mass (g). The mole of the molecule can be found on the periodic table, but first you need to know the molecule formula of the compound. Then add it up.
937526
If the reaction would "go" it is written: 3H2 + N2 = 2 (NH3)\nFill in the atomic masses of H = 1.00797 and N = 14.0067:\n[3 x 2 x (1.00797)] +[ 2 x (14.0067)]  = 2 x (14.0067) x 3 x (1.00797) = 34.06122\nThis is the stochiometric ratio of elements needed. You have 2.04 gm of H2, so multiply each term by 2.04 / 6.04782 to get:\n2.04 + 9.4492 = 11.4892. In other words, all of the H2 combines with about 9.45 gm of N2 to yield about 11.49 gm of ammonia with about 1.9 gm of N2 left over.
1097704
I believe it's a straight forward mole ratio conversion using the balanced equation.  For every one mole of Mg, one mole of H2 is produced.  \n\n.00164mol H2
433211
This is simple-\n3Mg(S) + N2(G) (2 is subscript) Yields (arrow) Mg3N2(S)
829148
I'll do one, and hopefully that'll help you figure out the rest.\n\n11. The first step is to create a balanced reaction.  I'm doing this from memory, and I could be wrong, so check it.\n2Zn + 2HCl ----> H2 + 2ZnCl\n\nNow figure out how many moles of zinc are in 3255 grams.  The atomic weight (aka the molar mass) of Zn is available from the periodic table:\n3255/65.409 = 49.764 moles\n\nSince you will have 2 moles of ZnCl for every 2 moles of Zn (a 1:1 ratio), you know that you will also end up with 49.764 moles of ZnCl.\n\nFigure out the molar mass of ZnCl by adding the atomic weight of Cl to that of Zn, then multiply this number by the total number of moles:\n(65.409 + 35.453)(49.764) = 5019.276g ZnCl\n\nUse the same method to solve all the rest.
655605
Apply the formula of\n\nM = molar strength wanted ( in M )\nW = molecular weight of the compound in question (in g/mol)\nV = the volume you wish to make (in mL)\nNo. = weight of compund needed (in grams)\n\napply to the formula\n\nNo. = (V * M * W)/1000 \n\nThen dissolve No. into the amoult of V you need (using a gradlated volumetric flask up to the volume you wanted)
979025
1. calculate the no. of mol of resulting MgO\n2. from the first eqn, 3 moles of MgO results from 1 mol of Mg3N2, find the no. of mol of Mg3N2\n3. then calculate grams of Mg3N2 (since all Mg3N2 must react with water)\n4. and find the percentage by grams (Mg3N2/0.470 g) x 100%
362155
Any case, the best way to calculate the yield of the product is from moles ratio.\nJust check howmany moles of reactants are participating and how many moles of product should form in theory... according to that you can get as idea about the yield of the product.\n\nif you can provide the reaction, i will help you more in detail\n\nGood luck!
851336
Since 1g=1mL, 1ppm=1mg/L.\n\n1. Convert ppm to mg/L\n\n2. Convert mg to g (divide by 1000)\n\n3. Convert g/L to moles/L (use the molecular weight)
856873
The balanced chemical equation and the molecular weights of the reactants and products.\n\nIn simplest form, assume\n\nXY + AB = XB + AY.\n\nAssume the molecular weight of XY is 50, and of XB 35.\n\nFor every 50 grams (or pounds, tons, or any other measure of mass) of XY you use, you will get 35 grams (or whatever) of XB.\n\nEDIT:\n\nOK. Atomic weights are H=1, S=32, Cl=35.5, Fe=56. Molecular weights are FeS=88, HCl=36.5, H2S=34, FeCl2=127.One gram molecular weight of each is its molecular weight in grams. So if you react 88 grams of FeS with 36.5 x 2 grams of HCl (the balanced equation tells you that you need 2 moles of HCl, so you multiply by 2), you will get 34 grams of H2S and 127 grams of FeCl2. \n\nNotice that the weights balance. You put in 88 + (36.5 X 2) grams, or 161 grams. You got out 127 + 34, or 161 grams.\n\nYou need the mass of any one of those compounds - either reactant or product - to determine all the others. Calculate the moles of the one for which you have the mass, determine the ratio of the moles of that one to the others, multiply by that ratio and then multiply result by molecular weight.