2.15g sample of KCL-KCLO3 mixture is decomposed by heating and produces 90.2cm3 O2(g).?
446321
Hmmm, a multi-step problem!  Love those!\n\nFirst, you will need to know how many moles of oxygen are generated.  For that, you will need to use the Ideal Gas Law:\n\nPV = nRT\n\nYou need to solve for n = PV/RT, and since you know the volume, pressure, and temperature, and the value of R is a constant, this is an easy calculation.\n\nThen, from the equation:  \n\nKClO3 ===> KCl + 1.5 O2\n\nyou can work backwards to determine the number of moles of KClO3 you started with (one mole of KClO3 generates one and a half moles of oxygen, thus the number of moles of KClO3 you started with will be 2/3 of the number of moles of oxygen generated).  Knowing how many moles of KClO3 you had, you can then determine the starting mass of the KClO3, then find out what percentage that was of your original starting mixture.\n\nGood Luck!
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
830588
If I understand your question, you start out with 2.15 g of AgNO3 and you isolate 1.16 g of AgCL.\n\n(It saves time if you include the M.Wt. of the materials.)\n\nAg = 107.87g/mole\nNO3- has 14.0067 + 3*15.9994 = 62.005 g/mole\nCL = 35.453 g/mole\nAgNO3 = 169.875 g/mole\nAgCL = 143.323 g/mole\n\nIn theory, for every mole of AgNO3 you should get 1 mole of AgCl\n\n(2.15 g AgNO3)*(1 mole AgNO3/169.875g) = 0.0126 mole AgNO3, and also the same number of moles of AgCL\n\n0.0126 mole AgCL * (143.323 g AgCl/mole AgCl) = 1.81 g AgCl in theory (rounded)\n\nYou isolated only 1.16 g, therefore\n\n  (1.16/1.81)*100% = 64 % yield.\n\nYou say you want % error, I don' t know if that is the correct term but here is what the error term is:\n\n [abs (1.16 - 1.81)/1.81]*100% = 35.9%\n\nplease check for errors
250762
2 HCl + CaCO3 ----> CaCl2 + H2O + CO2
603847
if its 2% by mass, thats 2 grams of KCl for every 100 grams of solution its in. \n\n2g KCl= 2g / (74.55g/mol)KCl = 0.0268mol KCl\n\nBecause there is one mole of Cl in KCl, this is also the number of moles of Cl in 100 grams of solution.\n\n0.0268mol Cl * (35.45g/mol)Cl = 0.951gCl/100gsolution\n\nThis is 0.951%\n\nnote: 35.45g/mol is the molar mass of Cl and 74.55g/mol is the molar mass of KCl. These were obtained from the periodic table.
952146
The first part of the question gives you the information you need to determine the "calorimeter constant", or the amount of heat required to change the temperature of the whole calorimeter set-up by 1 degree.\n\nThe calcium carbonate/HCl stuff is the reaction whose enthalpy change you are trying to determine, using the same calorimeter as before.  When you combine calcium carbonate and HCl, you'll get a reaction that produces CO2 and leaves you with a solution of calcium chloride.  So the overall reaction that you're looking at is:\n\nCaCO3 + 2 HCl --> CO2(g) + H2O + CaCl2\n\nCalcium carbonate will only decompose to CO2 and CaO when you heat it--not when you react it with an acid.\n\nHope this helps....I'll let you do the calculations.
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
849074
K2S + COCl2 --> 2KCl + COS
923888
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
116913
I think it is this:\n\n            \nC12H22O11(s) --- HNO3---->  12C(s) = 11H2O(g) + heat
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
726529
2*-393.5+3*-285.8+1366.8 = -277.6 kJ/mol
871536
Start with 2.948g K2Cr2O7. End with how many mL. K=39, Cr=52, O=16, K2Cr2O7=294. Multiply by 1molK2Cr2O7/294gK2Cr2O7. The gK2Cr2O7 cancel, leaving molK2Cr2O7. Multiply by 1000mLK2Cr2O7/0.5molK2Cr2O7. This comes from 0.5M, which means 0.5 mol in 1000mL. The molK2Cr2O7 cancel, leaving mL solution to be evaporated.\n\nStart with 45mL NaCl, end with gNaCl. Na=23, Cl=35.5, NaCl=58.5. Multiply by 0.125molNaCl/1000mLNaCl. This comes from 0.125M. The mLNaCl cancel, leaving molNaCl. Multiply by 58.5gNaCl/1molNaCl. The molNaCl cancel, leaving the gNaCl that is the answer.\n\nStart with 100g glucose, G. End with L of G. C=12, H=1, O=16, C6H12O6=180. Multiply by 1molG/180gG. The gG cancel, leaving molG. Multiply by 1L G/0.5molG. This comes from 0.5M G. molG cancel, leaving L of G. Which is the answer.
962694
Tehe answer is:\nAg + HNO3 -> AgNO3 + H\nAgNO3 + KBr -> AgBr + KNO3
433948
According to the MSDS for KCl,\nits solubility at 0 degrees C is 28.1 g/100 g of water
251863
At 20 degrees C, it's 0.6g/100mL. You can convert that to a Ksp.
532416
It is CS2+3(O2)=CO2+2(SO2)