Chemistry Help!! : Ok so Ga reacts with Cl. 50.70grams of Cl2 are used. How many grams of product do you form?
91712
83.93 grams of GaCl3 are produced.\nHow?\nWell, note the following:\n35.453 g Cl/mol Cl\n70.91 g cl2/mol Cl2\n69.723 g Ga/mol Ga\n\n1 mole of GaCl3 = 3(35.453 g Cl) + 1(69.723 g Ga) = \n176.082 g GaCl3\n\n(50.7 g Cl2)*(1 mol Cl2 / 70.91 g Cl2)*(2 mol Cl/1 mol Cl2)*(1 mol GaCl3 / 3 mol Cl)*(176.082 g GaCl3 / 1 mol GaCl3) = \n83.93 g GaCl3\n\nThis means that for every 3 moles of Cl used, there will be 1 mole of GaCl3 produced.  It also means that for every mole of Cl2 used, 2 moles of Cl are going into the product.  Therefore for every 3/2 moles of Cl2 used, 1 mole of GaCl3 is produced.
945679
When  excess of SO2 is present the amount of product will only be controlled by amount of Cl2 present.\nNow, 2 moles of  Cl2 forms 1 mole of Cl2O\nThereofre, 1 mole of Cl2 will form (1/2) i.e. 0.5 moles of Cl2O\nThus, 0.400 moles of Cl2 will form 0.200 moles of Cl2O.\n\nHence answer is C.
889632
The Ag in the product can not appear out of no where. I assume the equation should have been written:\n\nAl + AgNO3  >  Al(NO3)3 + Ag\n\nSo . . . Al + 3AgNO3  >  Al(NO3)3 + 3Ag\n\nGet an "A".
859821
2.2 gram\nu can calculate it by the molarity formulae
969093
H2 = 2g/mol   2.05g = 1.025mol\nCO = 28g/mol    15.85g = .57mol\n\n2H2 + CO --> CH3OH\n\nmust be reduced to \nH2 + (1/2)CO --> (1/2)CH3OH\n\n1.025mol + (1/2).57mol --> the quantity of the limiting reactant which is .285mol CH3OH which at 32g/mol = 9.12g
589036
The correct answer is 4 because on the reactant side you have 2 of K and 2 of Cl. On the product side you have 2K and 2 Cl. The reactant side and the product side are both balanced.
527028
This time i ans ur question but next time plz do ur homewrk ur self.\n\n1.\nWeight of NaOH = (4*.05) = .2 grams\nMoles of HCl =  Moles of NaOH = (.2/40) = .005\nVol. of HCl = .005*22.4 = .112 l = 112 ml\n\n2.\nWell let the grams be x\n1000x = 74.5*2.25*(75+x)\n(1000-167.25)x = 167.25*75\nx = 14.98 grams
876667
Fe + CuCl2 --> FeCl2 + Cu\n\nmole of CuCl2 = 8.00 / 134.5 = 59.47 mmole\n\nmole of Fe = 5.40 / 56 = 96.42 mmole\nmole of unreacted Fe = 4.08 / 56 = 72.85 mmole\nmole of reacted Fe = 96.42 - 72.85 = 23.56 mmole\n\nmole of unreacted CuCl2 = 59.47 - 23.56 = 35.09 mmole\n\nYep, there will be around 35 mmole CuCl2 left in the beaker.\n\npeace\nvixklen
746173
17.5 of what? gram ms, mg, kg, lb?\n\nby definition Molarity = atomic mass (gm)/1 liter\nMgCl2 – has (24.31 + (34.45)x2)=94.21\n\nso 1 mole of  MgCl2 = 94.21/1 liter\n\nSo since M=m/v\nassume  it is 17.5 gram ms in 2.5  Molar solution then \nv=m/(M x 2.5)=(17.5/94.21)/2.5=074=74 ml
592234
3g of hidrogen (plus 14 g of nitrogen) make up 17g of ammonia (H+H+H+N=1+1+1+14), so:\n\n3g  H____17g NH3\n50g H____=283.33g NH3\n\nI assumed a 100% efficient reaction (that means every bit of reactant will actually react).
342141
ok well, the equation is balanced so you need equal moles of Ag and Cl to make the products. They give you the concentration and volume of Ag, so you can find the amount of moles by multiplying 0.104mol/L * 0.0272L = 0.0028288 mols of Ag. so 0.0028288 mols of Cl were there originally. and you have the volume of water, so you divide moles by volume and you get the concentration. 0.0028288 mols/0.025L = 0.113152 M. \n\nanother way to do it is the easy way, use c1v1 = c2v2. you have c1 = 0.104M, v1 = 27.2mL, v2 = 25mL, so you can find c2. you dont have to worry about units in this case so its easier. \n\nhope this helped. let me know if u need more help. laterz
882018
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
261494
It depends on what Chlorine you mean. There's Cl(2) which is Chlorine gas and then there's the chlorine that you put into pools and such. Roughly about 40 million global tons is produced each year.
745115
The product will be MgCl2 because:\n***Chlorine has 7 valent electrons which is 1 electron away 8 stable electrons (octet) so it will try to steal 1 electron the other substances if these substances let it.\n***Mg has 2 valent electrons which either 2 electrons too many or 6 electrons away form the octet (stable form). Therefore. it either steals 6 more electrons or just simply give its 2 extra electrons away to achieve the octet. Since giving away 2 electrons is much easier than stealing 6 electrons, it gives away its 2 electrons to 2 chlorine atoms.\n***Ar has 8 valent electrons; it is already stable so it won't react any other substances including Cl2 and Mg.
133194
This usually happens in solution, the Chlorine is a chloride ion.\n\nAgNO3(aq) + Cl-(aq) --> AgCl(s) + NO3-(aq)
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%
942319
This is what you should be looking at:\n\nHCl(gas) + NH3(gas)  -->  NH4Cl (solid)