Respuesta :

kenmyna
The  theoretical  yield  is  calculated  as  follows


write  the   equation  for  reaction
that  is  2 Zn  +  I2  ------>  2 znI

then  find  the  moles  of Zn  used
moles=  mass/molar  mass
= 125/  65.4  = 1.911moles

by  use  of  mole  ratio  between  Zn  to  Zni  which  is  2:2  the  moles  of  ZnI   is  also=   1.911moles

theoretical  yield  is therefore=  moles  of ZnI  x molar  mass  of   znI


= 1.911  moles  x     319.2 g/mol =   609.99   grams



sebassandin

Answer:

[tex]m_{ZnI_2}^{theoretical}=610.3gZnI_2[/tex]

Explanation:

Hello,

In this case, the required chemical reaction is:

[tex]Zn(s)+I_2(g) \rightarrow ZnI_2(s)[/tex]

In such a way, if 125 g react with sufficient iodine in stoichiometric proportions (no excess), the theoretical yield of zinc iodide turns out as shown below:

[tex]m_{ZnI_2}^{theoretical}=125gZn*\frac{1molZn}{ 65.38gZn}*\frac{1molZnI_2}{1molZn}*\frac{319.22gZnI_2}{1molZnI_2} \\m_{ZnI_2}^{theoretical}=610.3gZnI_2[/tex]

Considering that the molar ratio zinc to zinc iodide is 1 to 1.

Best regards.

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