Question #b2388

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Question #b2388

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Answer 1 · 2014-12-27T13:26:11

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 ( $Z n I_{2}$ $ZnI_2$ ) 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

$Z n + I_{2} \to Z n I_{2}$ $Zn + I_2 -> ZnI_2$

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

$m_{Z n . r e a c t} + m_{I_{2}} = 30.3 g$ $m_(Zn.react) + m_(I_2) = 30.3g$ -> this is true only for what participates in the reaction, the mass of unreacted $Z n$ $Zn$ should not be included here (this is what $m_{Z n . r e a c t}$ $m_(Zn.react)$ symbolizes).

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

$m_{Z n I_{2}} = m_{Z n . r e a c t} + m_{I_{2}} = 30.3 g \to$ $m_(ZnI_2) = m_(Zn.react) + m_(I_2) = 30.3g ->$
$m_{Z n . e x c e s s} = 48.12 - 30.3 = 17.8 g \to$ $m_(Zn.excess) = 48.12 - 30.3 = 17.8g ->$ excess $Z n$ $Zn$ ;

Knowing that we have a $1 : 1$ $1:1$ mole ratio between $Z n$ $Zn$ and $I_{2}$ $I_2$ , and that their molar masses are $65.4 \frac{g}{m o l}$ $65.4g/(mol)$ and $253.8 \frac{g}{m o l}$ $253.8g/(mol)$ , respectively, we can determine how much of each actually reacts:

$n_{Z n . r e a c t} = n_{I_{2}} \to \frac{m_{Z n . r e a c t}}{65.4 \frac{g}{m o l}} = \frac{m_{I_{2}}}{253.8 \frac{g}{m o l}}$ $n_(Zn.react) = n_(I_2) -> m_(Zn.react)/(65.4g/(mol)) = m_(I_2)/(253.8g/(mol))$ , (1) and

$m_{Z n . r a c t} + m_{I_{2}} = 30.3 g$ $m_(Zn.ract) + m_(I_2) = 30.3g$ (2)

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

$m_{Z n . r e a c t} = 6.21 g$ $m_(Zn.react) = 6.21g$ and $m_{I_{2}} = 24.09 g$ $m_(I_2) = 24.09g$

Since the number of $Z n I_{2}$ $ZnI_2$ moles must equal the number of $Z n$ $Zn$ and $I_{2}$ $I_2$ moles as well, we can check the result by

$n_{Z n I_{2}} = \frac{30.3 g}{(65.4 + 253.8) \frac{g}{m o l}} = 0.0950$ $n_(ZnI_2) = (30.3g)/((65.4 + 253.8)g/(mol)) = 0.0950$ moles, which equals

$n_{Z n . r e a c t} = \frac{6.21 g}{65.4 \frac{g}{m o l}} = 0.0950$ $n_(Zn.react) = (6.21g)/(65.4g/(mol)) = 0.0950$ moles and
$n_{I_{2}} = \frac{24.09 g}{253.8 \frac{g}{m o l}} = 0.0950$ $n_(I_2) = (24.09g)/(253.8g/(mol)) = 0.0950$ 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 $Z n$ $Zn$ and 24.09 g of $I_{2}$ $I_2$ , 30.3 g of $Z n I_{2}$ $ZnI_2$ are produced $\to$ $->$ $I_{2}$ $I_2$ is the limiting reagent.

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Question #b2388

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