A theoretical yield is the amount of substance you calculate mathematically, as opposed to the actual yield, which is the amount of substance you find using a balance.
To find the theoretical yield, you need a balanced equation. In this case,
Zn + I2 -> ZnI2
So, that was easy enough. The theoretical yield is limited by the reagent present in the smaller quantity by moles, not by mass. You must take the mass you have of zinc multiplied by zinc's molar mass to find the number of moles of zinc. You must take the mass you have of I2 and multiply by the molar mass of I2.
The limiting reagent is the substance of fewer moles. Since these two reagents react in a 1:1 ratio, you just need to compare and see which you have less of. This will give you the number of moles you can make of ZnI2, theoretically -- the theoretical yield.
The chemical formula of zinc iodide is ZnI2.
Equation is Zn + Pb(NO3)2 --> Zn(NO3)2 + Pb
Balanced equation is: Zn(s) + I2(s) --> ZnI2(s) Pick any initial mass of Zn and I2, and convert those masses to moles: 100 g Zn / 63.4 g/mol = 1.58 moles 100 g I2 / 253.8 g/mol = 0.394 mol I2 Since these two react in a 1:1 ratio, you will have used 0.394 moles of Zn to react with the 0.394 moles of I2. That would leave 1.58 - 0.394 = 1.186 moles of Zn unreacted. So, the fraction of the original zinc remaining would be 1.186 / 1.58 = .75 by sonu gupta
This answer balances a theoretical reaction that does not occur in nature. 2KOH + Zn -> Zn(OH)2 + 2K Zn actually dissolves in strong base to form a so called "zincate" which was traditionally gven the formula of ZnO22- . It is now recognised that these zincate ions are hydrated to various degrees and one such ion is (Zn(OH)4)2- . Zn + 2KOHaq -> K2ZnO2aq +H2
No. Zn + 2HCl = ZnCl2 + H2
Zn + PbCl2 --> Pb + ZnCl2
Equation is Zn + Pb(NO3)2 --> Zn(NO3)2 + Pb
2AgNO3 + ZnI2 -> 2AgI + Zn(NO3)2
ZnO + C --> Zn + CO
Balanced equation is: Zn(s) + I2(s) --> ZnI2(s) Pick any initial mass of Zn and I2, and convert those masses to moles: 100 g Zn / 63.4 g/mol = 1.58 moles 100 g I2 / 253.8 g/mol = 0.394 mol I2 Since these two react in a 1:1 ratio, you will have used 0.394 moles of Zn to react with the 0.394 moles of I2. That would leave 1.58 - 0.394 = 1.186 moles of Zn unreacted. So, the fraction of the original zinc remaining would be 1.186 / 1.58 = .75 by sonu gupta
This answer balances a theoretical reaction that does not occur in nature. 2KOH + Zn -> Zn(OH)2 + 2K Zn actually dissolves in strong base to form a so called "zincate" which was traditionally gven the formula of ZnO22- . It is now recognised that these zincate ions are hydrated to various degrees and one such ion is (Zn(OH)4)2- . Zn + 2KOHaq -> K2ZnO2aq +H2
No. Zn + 2HCl = ZnCl2 + H2
Br2 + Zn ----> ZnBr2
Zn + PbCl2 --> Pb + ZnCl2
Zn + CuSO4 --> ZnSO4 + Cu
The net ionic equation is Zn + 2H+ --> Zn2+ + H2
2Zn(s) + 2HCL (aq) --> 2ZnCl + H2 (g) The reactants are zinc (Zn) and hydrochloric acid (HCl).
Na and Zn are correct because they are elements. Na is monovalent so NaI is correct. Zn is divalent and so ZnI is incorrect and should be ZnI2 . So if we have ZnI2 we need 2 Na to remove the I2 . So balanced it becomes: 2Na + ZnI2 ----> 2NaI + Zn