# Question #b2388

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 ($Z n {I}_{2}$) 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

$Z n + {I}_{2} \to Z n {I}_{2}$

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

${m}_{Z n . r e a c t} + {m}_{{I}_{2}} = 30.3 g$ -> this is true only for what participates in the reaction, the mass of unreacted $Z n$ should not be included here (this is what ${m}_{Z n . r e a c t}$ symbolizes).

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

${m}_{Z n {I}_{2}} = {m}_{Z n . r e a c t} + {m}_{{I}_{2}} = 30.3 g \to$
${m}_{Z n . e x c e s s} = 48.12 - 30.3 = 17.8 g \to$ excess $Z n$;

Knowing that we have a $1 : 1$ mole ratio between $Z n$ and ${I}_{2}$, and that their molar masses are $65.4 \frac{g}{m o l}$ and $253.8 \frac{g}{m o l}$, respectively, we can determine how much of each actually reacts:

${n}_{Z n . r e a c t} = {n}_{{I}_{2}} \to {m}_{Z n . r e a c t} / \left(65.4 \frac{g}{m o l}\right) = {m}_{{I}_{2}} / \left(253.8 \frac{g}{m o l}\right)$, (1) and

${m}_{Z n . r a c t} + {m}_{{I}_{2}} = 30.3 g$ (2)

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

${m}_{Z n . r e a c t} = 6.21 g$ and ${m}_{{I}_{2}} = 24.09 g$

Since the number of $Z n {I}_{2}$ moles must equal the number of $Z n$ and ${I}_{2}$ moles as well, we can check the result by

${n}_{Z n {I}_{2}} = \frac{30.3 g}{\left(65.4 + 253.8\right) \frac{g}{m o l}} = 0.0950$ moles, which equals

${n}_{Z n . r e a c t} = \frac{6.21 g}{65.4 \frac{g}{m o l}} = 0.0950$ moles and
${n}_{{I}_{2}} = \frac{24.09 g}{253.8 \frac{g}{m o l}} = 0.0950$ 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 $Z n$ and 24.09 g of ${I}_{2}$, 30.3 g of $Z n {I}_{2}$ are produced $\to$${I}_{2}$ is the limiting reagent.