Calculte the gross profit one could make in the manufacture of KNO3 from NaNO3 and KCl if 100% yield could be obtained and KNO3 could be sold for $185/ton and NaNO3 cost $55/ton and KCl cost $28/ton.
132519
NaNO3 + KCL = KNO3 + NaCl\n\nso you need the atomic weight of each element\ni.e.\nNa (23 g/mol)\nN (14 g/mol)\nO (16 g/mol) x3 = 48 g/mol\nor 85 g/mol for NaNO3\n\nK (39 g/mol)\nCl (35.5 g/mol)\nor 74.5 g/mol\n\nKNO3 is 39 + 14 + 48 = 101 g/mol\n\nyou need one mole of NaNO3 and one mole of KCL to make one mole of KNO3.  One ton is 1000 kg, or 1,000,000 g.\nSo 85 g/mole * 55 dollars/1,000,000g = 0.004675 dollars/mole\n74.5 g/mole * 28 dollars/1,000,000g = 0.002086 dollars/mole\nor added together, you spend 0.006761 dollars/mole to make KNO3\nNow, you can sell KNO3 for 185 dollars/1,000,000g, and\nKNO3 = 101 g/mol * 185 dollars/1,000,000g = .018685 $/mole,\n So ((0.018685-0.006761)/0.006761)*100% = 176% gross margin
485936
The MW of KClO3 is 122.55 (ref.), so 0.251 g KClO3 is .0020481 M. This will evolve 3/2 * .0020481 = .0030722 M O2 * 22.4 L/M = .068818 L O2 at STP. Correcting for temperature, pressure and partial pressure H2O, multiply by 297/273 * 760 / (721.8 - 22.4) and volume at given temp. and press. = 0.081354 L.
338927
IAnd for your other request, I would request you to send a seperate mail to me through yahoo Answer so that I am able to send it part by part.These are long subjects and would take some time,The balanced equation would be                                           H3PO4+3KOH=K3PO4+3H2O
431774
First write the balanced chemical equation:\n\nPb(NO3)2 + 2NaBr --->  PbBr2  +  2NaNO3 \n\nThen, find the total volume of the solution: 200mL + 200mL = 400mL \n\nFrom the density, the mass is:   density = mass/volume   mass = density * volume = 1.0 g/mL * 400 mL = 400 g\n\nQ = mC(Tf - Ti) ; where Q = heat released, m = mass of solution,, C = specific heat capacity, and (Tf - Ti) = change in temperature\n\nSo,  Q = 400g * 4.2 J/gK * (2.44 + 273)  (Since C is J/gKelvin, the temperature rise also has to be in Kelvin)\n\nQ = 462.5 KJoules\n\nHope this helps...
471669
AgNO3 + NaCl => AgCl(white ppt) + NaNO3\nAgNO3 + KI => AgI( yellow ppt) + KNO3\n\nthere is a change of color because AgI is produced
178855
divide 200g by the molecular weight of KCL and then divide that by .8
18255
3AgC2H3O2 + Na3PO4 -----> Ag3PO4 + 3NaC2H3O2
512631
This is the smelting of Iron (III) oxide to metalic Iron using coke gas.  The balanced equation is:\n\nFe2O3 (s) + 3 CO (g)  --->  2 Fe (l) + 3 CO2 (g)\n\nOne mole of oxide produces 2 moles of Iron.  If 5.90 moles of Iron is produced, then 5.90/2 moles of oxide is needed.\n\nThe moles needed times the molecular mass of the oxide will give you the grams of oxide needed.  The molecular mass of Iron (III) oxide is  159.69 g/mol.
453931
KOH + HNO3 ---> KNO3+ H20\n\nmoles of KOH = 0.025L*0.2mol/L = 0.005 mol\nconcentration of nitric acid solution = 0.005mol/0.030L = 0.167M\nconcentration of KNO3 = 0.005mol/0.055L = 0.091M
403482
a) 3SiO2 +2 Al --->3 Si + 2Al2O3\n\nb)AgNO3 +HCl---> AgCl +HNO3\n\nc) Mg + CuSO4 ---> MgSO4 + Cu\n\nd) Pb(NO3)4 +4 NaI --->4 NaNO3 + PbI4 (s)
363255
2 KClO3--->  KCl + 3O2 \n\nthe molecular weight of KClO3 is 122.5\n\nso you can see that 2*122.5 = 245g of KClO3 corresponds to 3 moles of O2\n\n0.261 g corresponds to 0.261*3/122.5 = 0.064 moles of O2\n\nto calculate the volume you use the  formula  pV = nRT ; \nT = 293 K  and 1mmHg = 133 Pa\n\nV = 0.064*8.31 * 293/ 722.7*133 =0.0016m^3 = 1.6 liter
183108
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
371080
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
102887
Assuming the reaction is elementary:\n\nK = [03]^2/[02]^3\n\n1.8*10^-7 = [03]^2/0.0012^3\n\nSolving for [03], O3 concentration is 1.76*10^-8 M.  (Actually plus or minus, but the negative concentration is meaningless).
62871
Okay to answer the stoichiometry problem first...You need to balance the equation first...If you don't know how to balance the equation email me and I will explain.  But the balanced equation is:\n\n2Na3PO4 + 3Ba(NO3)2 ----> Ba3(PO4)2 + 6NaNO3\n\nHere is a basic "map" you need to follow:\n\nGrams of Substance A divided by molar mass of A ===> Moles of A times the coefficients of A and B from balanced equation to get moles of substance B===> Moles of substance B times molar mass of substance B ===> Grams of substance B\n\n\nTo start we need to calculate the molar mass of the two compounds we are mixing together\nNa3PO4 = 151.97g/mol\nBa(NO3)2 - 213.34g/mol\nThen we need to find the molar mass of the product, which is Ba3(PO4)2 - 577.97g/mol\nIf you don't know how to find the molar mass, email me. Otherwise, read a chemistry book.  Anyway the steps are to convert grams to moles then multiply by the coefficients of our balanced equation and then multiply again by grams of \n Ba3(PO4)2...Seems complicated, but it's really not.  \n\n1. To see if Na3PO4 is the limiting reactant we divide 3.50g by it's molar mass of 151.97g/mol to get moles of Na3PO4\n\n3.50/151.97 = 2.30*10^-2 mol. Take this and multiply by the coefficients of our reactant and product:\n\nTake the coefficient in front of Na3PO4 and place that in the denominator  /2 \nTake the coefficient in front of our product Ba3(PO4)2 and put it in the numerator  1/\n\nNow take the calculated moles of \n2.30*10^-2 and multiply them by (1/2) <----  the coefficients of reactant and product\n\nThis gives you moles of Ba3(PO4)2\n(2.30*10^-2) * (1/2) = 1.15*10^-2 mol\nNow multiply this number by the mol weight of our product which is 577.97g/mol and that gives you grams of product produced.\n\n(1.15*10^-2) * 577.97 = 6.66g of Ba3(PO4)2\n\nWhew!\n\nNow on to the next one....I won't explain fully as it's exactly the same procedure.\n\n6.40g/213.34g/mol = 2.99*10^-2 mol of Ba(NO3)2 (reactant)\n\n(2.99*10^-2) * (1/3) = 9.99*10^-3 mol of Ba(PO4)2 (product) \nI used (1/3) because the number in front of the reactant is 3 (goes in the denominator) and the 1 is the coefficient in front of the product.\n\nNow   (9.99*10^-3) * (577.97) = 5.78g of Ba3(PO4)2\n\nokay so upon inspection, the limiting reactant is Ba(NO3)2 and it yields 5.78 grams...\n\nYou ask if 3.50 grams are actually yielded, what is the %\n\nSo....(actual yield/theoretical yield)*100 = % \n\n(3.50/5.78)*100 = 60.55%\n\nAlrighty....Now onto problem #2\n\nBalance: Al(NO3)3 + Na2S ----> Al2S3 + NaNO3      \n\n(A) Al(NO3)3 + (B) Na2S ----> (C) Al2S3 + (D) NaNO3\n\nPut variables in front of each reactant and product like I did above.  Now assign an element to equal the variable on both sides.  In our case:\n\nAl  -    A = 2C\nNO3 -   3A = D\nNa  -   2B = D\nS   -   B = 3C\n\nNow make one letter equal one. We'll make A = 1\nSo..\nA = 1\nB = 3/2  (by algebra)\nC = 1/2  (by algebra)  (email me if you don't know)\nD = 3\n\nwe can't have fractions in there, so we multiply all of them by two to get the 2 in the denominators out, and we end up with\n\nA = 2\nB = 3\nC = 1\nD = 6    With a final equation balanced out to look like:\n\n2Al(NO3)3 + 3Na2S ---> Al2S3 + 6NaNO3\n\nHope all this was clear!!!  Enjoy
452368
Oh, that Gopal.\n\ndP/dt requires multiplication, not division.\n\ndP/dt = (dP/dx) * (dx/dt)\n\ndP/dx is as Gopal gives it,\n\ndP/dx = -0.8x+24\n\ndx/dt is given as 6. When x = 25, we have\n\ndP/dx = -.8(25) + 24 = 4\n\nThen\n\ndP/dt = 4 * 6 = $24 per day, or $1 per hour\n\nNotice that profit per unit declines as production increases. Each unit yields .8 less profit than the previous. (The slope of dP/dx is negative.)\n\nAs a rough check, the profit per day when 25 units per day are produced is\n\nP = -.4x^2 + 24x - 10\n = -.4(625) + 24(25) - 10\n = -250 + 500 -10\n = 340\n\nFour hours later, the rate of production is 26 units per day and the profit per day is\n\nP = -.4(26)^2 + 24(26) - 10\n = 343.6\n\nThe profit is not the full 344 per day because the profit per unit falls off as the number of units increases. Profit is increasing at the rate of $1 per hour only at the instant given in the problem, and then falls off as production increases.
535327
Hi. As with most things it depends on the cost of manufacture. : http://www.steelonthenet.com/commodity_prices.html