Question

Determine the limiting reactant in a mixture containing 139 g of Se, 431 g of Cl2, and 110 g of H2O. Calculate the maximum mass (in grams) of selenic acid, H2SeO4, that can be produced in the reaction.

Answer 1

Answer : The limiting reagent is $H_2O$

The mass of $H_2SeO_4$ produced is, 220.4 grams.

Explanation : Given,

Mass of Se = 139 g

Mass of $Cl_2$ = 431 g

Mass of $H_2O$ = 110 g

Molar mass of Se = 79 g/mol

Molar mass of $Cl_2$ = 71 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of Se, $Cl_2$ and $H_2O$.

$\text{Moles of }Se=\frac{\text{Given mass }Se}{\text{Molar mass }Se}$

$\text{Moles of }Se=\frac{139g}{79g/mol}=1.76mol$

and,

$\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}$

$\text{Moles of }Cl_2=\frac{431g}{71g/mol}=6.07mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{110g}{18g/mol}=6.11mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Se(s)+3Cl_2(g)+4H_2O(l)\rightarrow H_2SeO_4(aq)+6HCl(aq)$

From the balanced reaction we conclude that:

As, 4 mole of $H_2O$ react with 1 mole of $Se$

So, 6.11 moles of $HCl$ react with $\frac{6.11}{4}=1.52$ moles of $Se$

and,

As, 4 mole of $H_2O$ react with 3 mole of $Cl_2$

So, 6.11 moles of $HCl$ react with $\frac{6.11}{4}\times 3=4.58$ moles of $Cl_2$

From this we conclude that, $Cl_2\text{ and }Se$ are excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2SeO_4$

From the reaction, we conclude that

As, 4 mole of $H_2O$ react to give 1 mole of $H_2SeO_4$

So, 6.11 mole of $H_2O$ react to give $\frac{6.11}{4}=1.52$ mole of $H_2SeO_4$

Now we have to calculate the mass of $H_2SeO_4$

$\text{ Mass of }H_2SeO_4=\text{ Moles of }H_2SeO_4\times \text{ Molar mass of }H_2SeO_4$

Molar mass of $H_2SeO_4$ = 145 g/mole

$\text{ Mass of }H_2SeO_4=(1.52moles)\times (145g/mole)=220.4g$

Therefore, the mass of $H_2SeO_4$ produced is, 220.4 grams.