Question #b2388

1 Answer
Dec 27, 2014

First of all, I think the problem was written incorrectly - I am referring to the value given for the total mass of the reactants, which I believe was 30.3 g, not 3.03 g, and to the fact that the product is zinc iodide (#ZnI_2#) Therefore, I'll try and make this more of a concept-problem.

So, according to the law of conservation of mass, the total mass of the reactants must be equal to the total mass of the product - mass can neither be created, nor destroyed in any ordinary chemical reaction.

Let's assume that this is the reaction we're looking for

#Zn + I_2 -> ZnI_2#

Since the total mass of the products is #30.3g#, we know that

#m_(Zn.react) + m_(I_2) = 30.3g# -> this is true only for what participates in the reaction, the mass of unreacted #Zn# should not be included here (this is what #m_(Zn.react)# symbolizes).

We know that the mass of the product must equal the mass of the reactants, so

#m_(ZnI_2) = m_(Zn.react) + m_(I_2) = 30.3g -> #
#m_(Zn.excess) = 48.12 - 30.3 = 17.8g -># excess #Zn#;

Knowing that we have a #1:1# mole ratio between #Zn# and #I_2#, and that their molar masses are #65.4g/(mol)# and #253.8g/(mol)#, respectively, we can determine how much of each actually reacts:

#n_(Zn.react) = n_(I_2) -> m_(Zn.react)/(65.4g/(mol)) = m_(I_2)/(253.8g/(mol))#, (1) and

#m_(Zn.ract) + m_(I_2) = 30.3g # (2)

I won't detail the solving of this equation system because it's too simple; solving for the two masses will show that

#m_(Zn.react) = 6.21g# and #m_(I_2) = 24.09g#

Since the number of #ZnI_2# moles must equal the number of #Zn# and #I_2# moles as well, we can check the result by

#n_(ZnI_2) = (30.3g)/((65.4 + 253.8)g/(mol)) = 0.0950# moles, which equals

#n_(Zn.react) = (6.21g)/(65.4g/(mol)) = 0.0950# moles and
#n_(I_2) = (24.09g)/(253.8g/(mol)) = 0.0950# moles.

As a conclusion, the results of this reaction agree with the law of conservation of mass: from a total of 6.21 + 17.8 = 24.01 g of #Zn# and 24.09 g of #I_2#, 30.3 g of #ZnI_2# are produced #->##I_2# is the limiting reagent.