Question ae352

May 14, 2017

You need 143 g of ${\text{KClO}}_{3}$.

Explanation:

Step 1. Calculate the moles of $\text{CO}$

We can use the Ideal Gas Law to calculate the number of moles.

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

NTP is 20 °C and 1 atm.

n = (pV)/(RT) = (1.0 cancel("atm") × 84 cancel("L"))/("0.082 06" cancel("L·atm·K⁻¹")"mol"^-1 × 293.15 cancel("K")) = "3.49 mol"#

Step 2. Calculate the moles of ${\text{O}}_{2}$

The balanced equation is

${\text{2CO + O"_2 → "2CO}}_{2}$

${\text{Moles of O"_2 = 3.49 color(red)(cancel(color(black)("mol CO"))) × "1 mol O"_2/(2 "mol CO") = "1.75 mol O}}_{2}$

Step 3. Calculate the mass of ${\text{KClO}}_{3}$

The balanced equation is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m l} 122.55$
$\textcolor{w h i t e}{m m m} {\text{2KClO"_3 → "2KCl" + "3O}}_{2}$

${\text{Mass of KClO"_3 = 1.75 color(red)(cancel(color(black)("mol O"_2))) × (2 color(red)(cancel(color(black)("mol KClO"_3))))/(3 color(red)(cancel(color(black)("mol O"_2)))) × "122.55 g KClO"_3/(1 color(red)(cancel(color(black)("mol KClO"_3)))) = "143 g KClO}}_{3}$