CuSO4 + H2O + Al + NaCl Yields...?
471526
I suspect that the product is AlSO4.\n\nI'm thinking that the Al swaps with the Cu.  That would leave a free Cu+ which might displace the H from H20 and form either CuOH or CuO, either one of which would probably produce H2.  \n\nSo, I guess you would first get CuOH and AlOH, as intermediate reactions.\n\nI think the equation is:\n\n                          NaCl (goes over the arrow)\n6CuSO4 + 4Al + 12H2O ---------------> 2Al2(SO4)3 + 6Cu(OH)2 + 6H2
745889
We can solve this using a system of equations.\n\nLet x be the volume of the 25% solution. Let y be the volume of the 60% solution.\n\nThe sums of the volumes must total 140 ml:\n\nx + y = 140.\n\nThe mass of CuS04 in the desired solution must equal the amounts added through the use of both solutions.\n\n30 percent of 140ml gives 42 ml of "pure" copper sulfate solution.\n\nThe 25% solution would contribute .25x of "pure" copper sulfate solution. The 60% solution would contribute .60y of "pure" copper sulfate solution, so:\n\n0.25x + 0.60y = 42.\n\nHere is the system:\n\nx + y = 140\n0.25x + 0.60y = 42\n\nThere are many different ways to solve the system. I'll use substitution here:\n\nx + y = 140, so y = 140 - x\n\n0.25x + 0.6(140 - x) = 42\n0.25x + 84 - 0.6x = 42\n-0.35x = -42\nx = 120\ny = 20\n\nSince x is the amount of 25% solution and y is the amount of 60% solution, we need 120 ml of 25% solution and 20 ml of 60% solution.\n\nThis appears to match your answer!
872094
A . 4LiH + AlCl3 → 2LiAlH4 + 2LiCl <-- there's only 1 alluminum so it can't yield 4 in the product\n\nB. Sn + 2H2SO4 → SnSO4 + SO2 + 2H2O <-- there are 1 sn, 4 H, 2 S and 8 O ( multiply the coefficient [2] by the subscript [4])\nit matches in the product\nCORRECT\n\nC. 4BF3 + 3H2O → H3BO3 + 3HBF4< same goes here, just write out how many of each element are in the beginning of the equation and in the product and compare\nCORRECT\n\nD. 3Cl2 + 6NaOH → 5NaCl + NaClO3 + 3H2O <-- this one is hard to read, I'm assuming that I is not an L because it's followed by a subscript, so it would be iodine\nbut either way, this one is wrong because there are 5 C/Cl in the product which is wrong
814611
CdSo4+NH4OH ---->Cd(OH)2 + (NH4)2SO4
745644
H3PO4 + 3NaOH ----> 3H2O + Na3PO4
890642
Al + Fe(NO3)3 ---> Al(NO3)3 + Fe.\n\n2K + 2H2O ---> 2KOH + H2\n\n2KOH + CuSO4 ---> Cu(OH)2 + K2SO4\n\n3NaOH + H3PO4 ---> 3H2O + Na3PO4\n\nBest?
969567
You can evaporate, or distill off the water.  Then you'll get the CuSO4, along with any non-volatile impurities that may be present.  Or you can do a recrystalization.  You boil the solution until the CuSO4 just starts to crystalize out.  Then you let the solution cool.  As it cools, the solubility of the salt decreases, and more & more of it crystalizes.  When cooling, you may end up with a super-saturated solution, instead of crystals being formed.  In that case, scratching the inside of the beaker with a glass rod will initiate the crystalization process.  With recrystalization, the product is also purified.
1097630
Na2CO3 + 2HCl-> NaCl + H2O + CO2
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
223710
Start by balancing Al, then Cl. Al is already  balanced, so when you balance Cl everything is easy from their. \nYou final amswer will be:\n3HCl + Al(OH)3 ----> 3H2O + AlCl3
70269
Cu+S+O*4+H*2+O\n63.55+32.07+16*4+1.008*2+16= 177.64 g/mol\nThat x (usually a dot) means that there is a water molecule attached to the CuSO4.  The water molecule's mass should be included in the molar mass calculation.
429906
Never trust materials science majors ...\n\n1. NaCl (s) --> Na+ (aq) + Cl- (aq)\n2. Ca (s) + 2H2O (l) --> Ca(OH)2 (s) + H2 (g)\n3. Ca(OH)2 (s) --> H2O (g) + CaO (s)
782516
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)
1184650
According to the equation\n\n2 AlCl3 + 3 Na2S ---------Al2S3 +6 NaCl\n\n2 moles of Aluminium Chloride can produce one mole of aluminium sulphide.\nOne mole can produce 1/2 moles of aluminium sulphide.\n\n3.5 moles of aluminium chloride can produce 1/2 X3.5 moles =1.75 moles of aluminium sulphide.
866145
HCl + AgNO3-----> AgCl +HNO3\nPb(OOC-CH3)2  + 2LiCl -------> PbCl2  + 2Li(OOC-CH3)\nNa3PO4 + CrCl3 -------> Cr(PO4) + 3NaCl\nH2SO4 + BaCl2 ----------> BaSO4 + 2HCl\nAl2(SO4)3  + 6NH4OH ------> 2Al(OH)3  + 3(NH4)2SO4\n2AgNO3 + H2S -------> Ag2S  + 2HNO3\nCaCl2   + Pb(NO3)2 -------> Ca(NO3)2  + PbCl2\nCa(NO3)2 + Na2CO3 ------>CaCO3  + 2 NaNO3
51552
AgNO3 + NaCl -----> AgCl + NaNO3\n\ntotal ionic equation\n\nAg+  +  NO3-  +  Na+  +  Cl-\n\n                      ----->\n\n                            AgCl  +  Na+  +  NO3-\n\nnet ionic equation\n\n    Ag+  +  Cl-  ----->  AgCl
564230
This is a selective way to produce cupper of cupper hydroxides by leaching hydroxide and elecrtowinnig as:\nCu(OH)2(s) = Cu+2(aq) + 2 OH–(aq) \nCu+2(aq) + 4 NH3(aq) = Cu(NH3)4+2(aq)\n\nand [Cu(NH3)4]2+ is better agent than [Cu(H2O)6]2+ in (aq) solutions for this.