# What amount of Zn will be required to produce H_2 by its action on dilute H_2SO_4 which will completely react with the oxygen, produced by heating 30g of KClO_3?

Jun 25, 2018
• n("Zn")=0.651 color(white)(l) "mol"
• $m \left(\text{Zn") = n("Zn") * M("Zn}\right) = 42.56 \textcolor{w h i t e}{l} g$

#### Explanation:

$m \left({\text{KClO}}_{3}\right) = 30 \textcolor{w h i t e}{l} g$

n("KClO"_3) = 30 color(white)(l) g * (1 color(white)(l) "mol")/(138.55 color(white)(l) "g")=0.217 color(white)(l) "mol"

The balanced equation of the decomposition of potassium perchlorate

$\textcolor{n a v y}{2} \textcolor{w h i t e}{l} {\text{KClO"_3 (s) stackrel(Delta)(to) 2 color(white)(l) "KCl" (s) + color(purple)(3) color(white)(l) "O}}_{2} \left(g\right)$

suggests the stoichiometric relationship

• $\textcolor{n a v y}{2} \textcolor{w h i t e}{l} {\text{mol" color(white)(l) "KClO}}_{3}$ decomposes to produce $\textcolor{p u r p \le}{3} \textcolor{w h i t e}{l} {\text{mol" color(white)(l) "O}}_{2} \left(g\right)$

Hence the amount of oxygen required for the complete combustion of the unknown amount of ${\text{H}}_{2} \left(g\right)$ produced would be

n("O"_2) = 0.217 color(white)(l) "mol" color(white)(l) "KClO"_3 * (color(purple)(3) color(white)(l) "mol" color(white)(l) "O"_2 )/(color(navy)(2) color(white)(l) "mol" color(white)(l) "KClO"_3)=0.326 color(white)(l) "mol" color(white)(l) "O"_2

Oxygen reacts with hydrogen by the equation

$\textcolor{p u r p \le}{1} \textcolor{w h i t e}{l} \text{O"_2 (g) + color(navy)(2) color(white)(l) "H"_2 (g)stackrel("*")(to) 2 color(white)(l) "H"_2"O} \left(g\right)$

at a $\textcolor{p u r p \le}{1} : \textcolor{n a v y}{2}$ ratio, meaning that the combustion would consume

n("H"_2) = 0.326 color(white)(l) "mol" color(white)(l) "O"_2 * (color(navy)(2) color(white)(l) "H"_2 )/(color(purple)(1) color(white)(l) "O"_2)=0.651 color(white)(l) "mol" color(white)(l) "H"_2

of hydrogen. All these ${\text{H}}_{2}$ came from the reaction between $\text{Zn}$ and dilute sulfuric acid ${\text{H"_2 "SO}}_{4}$ as seen in the following equation

$\textcolor{n a v y}{1} \textcolor{w h i t e}{l} {\text{Zn"(s) + "H"_2"SO"_4 (aq) to color(navy)(1) color(white)(l) "H"_2 (g) + "ZnSO}}_{4} \left(a q\right)$

where for each mole of ${\text{H}}_{2} \left(g\right)$ produced, $\textcolor{n a v y}{1} \textcolor{w h i t e}{l} \text{mol}$ of $\text{Zn}$ is consumed. That is:

n("Zn") = n("H"_2) = 0.651 color(white)(l) "mol"

Hence the mass of $\text{Zn}$

$m \left(\text{Zn") = n("Zn") * M("Zn}\right) = 42.56 \textcolor{w h i t e}{l} g$