In the chemical equation Zn + 2HCL -- ZnCl² + H², the reactants are?
547048
D..zinc and Hdrochloric acid\n\nand zn replaces hdrogen because zinc iz more active that H in chamical activity series
844424
In a single replacement reaction, an uncombined element (such as zinc, Zn) reacts with a compound (e.g., hydrochloric acid, HCl), taking the place of one of the elements in the compound (hydrogen, H, in this case). the result is a compound that includes the formerly uncombined element (zinc chloride, ZnCl2) and an element that was part of the former compound but got replaced (hydrogen, which is now in gas form, H2).\n\nTo balance the equation, you need to Cl's for each Zn, so you put one Zn and 2 HCls on the left and see what comes out on the right:\n\nZn + 2 HCl  ->  ZnCl2 + H2\n\nZinc replaced hydrogen. Single replacement reaction.\n\nIn a double replacement reaction, you have two compounds, and the elements "change partners."\nExample: Zinc Carbonate (ZnCO3) and Hydrochloric Acid (HCl)\n\nThe most powerful oxidizing agent in the combination (Cl) will combine with the most powerful reducing agent (Zn). This leaves the less powerful oxidizing agent (carbonate, CO3) to combine with the less powerful reducing agent (H). Again, we will need to Cls for each Zn, so the equation balances as follows:\n\nZnCO3 + 2 HCl  ->  ZnCl2 + H2CO3\n\nZinc has replaced hydrogen, and hydrogen has replaced zinc.\nAlso, chlorine has replaced carbonate, and carbonate has replaced chlorine.\n\nResult: double replacement.
725081
The answer is Cu + Cu(NO3)2 -> Cu + Cu(NO3)2\n\nThere is no reaction.
737051
2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n\n39.9 mL x L / 1000 mL x .466 mol / L Ca(OH)2 x 2 mol HCl / mol Ca(OH)2 x L / .127 mol x 1000 mL / L = 292.81 mL HCl
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
801118
CuCl2 + 2NH4OH --> Cu(OH)2 + 2NH4Cl\n\nCu(OH)2 (s) + 4NH3 (aq)     ----->     Cu(NH3)4^2+
215729
K2IC4H6 + HNO3 ---  K2HI + NO3C4H6
954955
HCl_aq + NaOH_aq -> NaCl_aq + H2O_l\n(H+)_aq + (Cl-)_aq + (Na+)_aq + (OH-)_aq -> (Na+)_aq + (Cl-)_aq + H2O_l\n(H+)_aq + (OH-)_aq -> H2O_l\n\nFor all the other equations, the (Cl-), (K+), (NO3-), (Na+) are spectator ions, and the n.i.e is the same:\n(H+)_aq + (OH-)_aq -> H2O_l
1093944
i) add NaNH2(1 eq.)  then CH3I(1 eq.) (I=iodine)then NaNH2(1 eq.)  then C2H5I then add hydrogen in presence of palladium and BaSO4.\nii)add NaNH2(1 eq.) then C3H7I(1 eq.)  then add NaNH2(1 eq.)  then again C3H7I(1 eq.) then add ZnBr2 or MgBr2.
934206
2\nthe reaction is\nCa(OH)2+2HNO3=2H2O+Ca(NO3)2
127210
Hydrochloric acid will react with metals ABOVE Hyrogen in the activity series of metals to produce Hydrogen gas.\nThe metals will displace Hydrogen from the acid and make a metal salt (which is usually in solution).\n\nSo if we say that [A] is a generic metal above Hydrogen in the activity series that has a normal oxidation state of +1, then the reaction would be as follows,\n[A] (s) + HCl (aq) --> H2 (g) + [A]Cl (aq)
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
750179
NaCl2 + NaCO3 ==> NaCO3 + NaCl2 NOTE: these two reactants will not react because the reactants and the products are identical\n\nCu + O2 ==> CuO2   This is already balenced\nCuSO3 + Fe ==> FeSO3 + Cu   The charge is not given so procuts could be different\n\n2HCl + 2Zn ==> 2ZnCl + H2\nMnO2 + H2O2 ==> No reaction: similar to NaCl2+NaCo3\n\nGood luck
867182
6HCl +3 I2 + KClO3 --->6 ICl+ KCl + 3H2O\n\n\nMnO4 - +2 NO2- +4 H+ ---> Mn2+2 NO3- +2 H2O
808333
You have a common ion effect problem\n\nPbCl2 <=> Pb+2 +2Cl-\n\nKsp=[Pb+2][Cl-]^2\nYou know that [Pb+2]=5.5*10^-3\nAlso from the reaction you know that the amount of Cl- coming from the PbCl2 is double the amount of Pb+2.\nSo if C is the concentration of Cl- coming from HCl, the amount of Cl- at equilibrium is [Cl-]=C+2*[Pb+2] \n\nSo Ksp= [Pb+2]* (C+2*[Pb+2])^2=>\nif we assume that C>>2[Pb+2] then C+2*[Pb+2] =C and \nKsp=[Pb+2]C^2 =>1.7*10^-5= 5.5*10^-3 C^2 =>\nC=55.6 * 10^-3 but 2*[Pb+2]=11*10^-3, so we can't do the approximation...\n\nLet's name [Pb+2]=p for simplicity\n\nKsp=p(C+2p)^2 =>\nKsp/p= C^2+4pC+4p^2 =>\nC^2+4pC+4p^2-Ksp/p=0 =>\nC^2+2.2*10^-3 C -2.97*10^-3 =0\n\nC=0.5*(-2.2*10^-3 + SQRT( (2.2*10^-3)^2 +4*2.97*10^-3) ) =>\nC= 5.34*10^-2\n\nIf you added Va L of HCl, then\nC=mole HCl/Vtotal = Ca*Va/(Va+0.12) =>\nCVa+0.12C=CaVa =>\nVa=0.12C/(Ca-C)= 0.12*0.0534/(0.12-0.0534) = 0.0962L or 96.2 ml
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