What volume of hydrogen at stp is produced when 2.5 g if zubc react with an excess of Hydrochloric acid?
407430
n of Zn~ 2,5/65,4= 0,038 mole\nso, number of moles of H 0.076 (1 mol H20 has 2 mols H)\nPV=nRT\n1*V=0.076*22.4\nV = 1.70 L
655398
Here's how to solve it (I am convinced that you should do your own homework, but hints are OK as long as you learn something):\n\nStart solving by finding the molecular mass of the compound.\nA. Volume occupied by one mol of gas is 22.4 L at standard temperature and pressure (STP: 760 mmHg and 273 K (0 C))\nTo find the volume of gas that weighs 2.57 g at STP, use formula: P1V1/T1=P2V2/T2, where R is 0.08206, P1=319 mmHg, V1=2 L, T1=298 K; P2=760 mmHg, T2=273 K and V2 is unknown (it will be somewhat less than 1 L).\nAfter you get this volume, see how many mols do you have in it: divide 22.4 by V2 you just got. Then multiply this number by 2.57 and get the molecular mass of your compound (M(X)).\n\n2. Multipy M(X) separately by 0.6481, 0.136 and 0.2159 to get number of atomic weight units that are allocated to C, H, and O respectively. \nThen divide first number by 12, 2nd by 1 and 3rd by 16 to see how many atoms it amounts to. The resulting integer numbers are X, Y, Z.\n\n3. CxHyOz is the empirical formula of your compound. Google it to get the possible molecular ones. This is a trick question, so there will be more than one but less than 8. All of them are alcohols or ethers.
406601
When performing a chemical reaction you want to use as little of each material to generate the quantity of material you need, with as little waste as possible.\n\nFor example:\n\nHCl + NaOH -> NaCl + H2O\n\nYou are using 1 mole of HCl and 1 mole of NaOH to generate 1 mole of salt and 1 mole of water, therefore stoichiometryt is 1:1.  This is the stoichiometry.  \n\nThe reaction:\n\nH2SO4 + 2NaOH -> Na2SO4 + 2H2O\n\nThe stoiciometry is 1 mole of H2SO4 and 2 moles of NaOH, therefore 1:2\n\nStoichiometry is the relative amounts of each reagent you use in a reaction.\n\nIf you were to use:\n\n2HCl + NaOH -> NaCl + H2O + HCl\n\nYou are using 2 moles of HCl and 1 mole of NaOH to generate the same thing, but you have an excess of HCl left over.  This is yield.  The yield of this experiment in relation to HCl would be 50%, since you have only consumed half of it, the rest is waste.  With respect to NaOH the yield is 100% since you have used all of it.  The yield of the reaction though is 100% as it is always based on the limiting reagent.\n\nThe product yield is the amount of product you recover based on a theoretical maximum.  In the first example above, and assuming you did the reaction on a 1 mole scale, you should recover 1 mole of NaCl (57.5g) and 1 mole of H2O (18g).  If you only recovered 30g of NaCl, then the yield of the reaction is 30 / 57.5 = 52.2%.\n\nNo chemical reaction is perfect so getting 100% return is rarely possible, although we do get very close occasionally in industry.\n\nEcologically the less EXCESS of reagent you can put into a reaction to maximise the recovery limits the amount of material going to waste, which has an impact on the environment.  Economically the less EXCESS of reagent you need, means it costs less to prepare a compound.\n\nWhen pharmaceutical companies make drugs they spend millions on materials.  Cutting down by 10% how much of everything they need can save hundreds of thousands of pounds per year / per drug on excess materials expenditure, increased environmental exposure etc.  Chemical companies spend 100's of thousands of pounds a year disposing of excess chemicals safely, recycling chemicals etc and are liable to very heavy fines if they don't.  Many savings can be made just by improving stoichiometries and product yields.\n\nLast year I spent over $1 million on chemicals, by improving stoichiometries and yields by 10% this alone saves an enormous amount of money.\n\nI hope this helps?
917456
With hydrochloric acid the equation is:\n\n16HCl  +  2KMnO4  =  2KCl  +  2MnCl2  +  8H2O  +  5Cl2\n\nand with sulphuric acid :-\n\n2KMnO4  +  3H2SO4  =  K2SO4  +  2MnSO4  +  O2
385149
density of H2 at STP = d= PM /RT\n\nd= density  =X\n\nP= pressure  =1atm\n\nM= molar mass = 2.016g of H2\n\nR= gas constant = .0821\n\nT=temperature  = 273K\n\n\nX= .0899g/L\n\n\n 1.6g H2 corrected to STP: \n\n\n1.6gH2(.0899g/L / 2.016gH2)= .0713g/L
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
591191
if i remember my chemistry right, a mole of gas at stp is 22.4 liters, so the answer should be E), 2 moles
24431
3225..............x\nZn   +  2HCl -> ZnCl2 + H2\n65.4............136.4\n\nx=3225*136.4/65.4\nx= 6 726.14679 grams
868179
2Rb + 2 H20 --> 2RbOH + H2
938236
Step one: find the moles of H2SO4.  (2.7 moles per liter)*0.025 liters.\n\nStep two: multiply by two because the reaction is \n2NaOH + H2SO4 ----> 2H2O  + Na2SO4 and so you need two NaOH molecules per molecule of H2SO4.\n\nStep three: multiply by the molecular mass of NaOH.\n\nThat will get you the answer.
290781
MgSO4(s) + 2HCL(aq) ---> H2SO4(aq) + MgCl2 (aq)
511319
NaOH is in excess\n\nSubtract moles of H+ from OH- (.0345 mol OH- - .025 mol H+ = .0095 mol OH-)\n\nDivide by total volume (.0095 mol OH-/.0595 L = 0.1596 M)\n\nFind pOH = -log [OH-] = -log [.1596] = 0.7967\n\nFind pH = 14-.7967 = 13.20
1116694
the reaction leads to production of hydrogen gas ...\n\n\n2Al(s) + 6HCl(aq) → 2Al3+(aq) + 6Cl-(aq) + 3H2(g)
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.
35280
The Ksp of the reaction\n\nHgS <--> Hg + S\n\nis 2*10^-53. Extremely insoluble.
455269
H2SO4 + Mg(OH)2 -> 2H2O + MgSO4
878297
Assuming 1 mole (28.01348 g) N2 = 22.4 liters at STP, I get (28.01348*.00220102/22.4) =  0.0027526 g if collected at STP. Correcting by 273/293 * (1.5*760-17.5)/760 (for temp., press. and water) = 0.003788 g. \nSo I basically agree with your results. What's the ans. supposed to be?