Question #ae352

1 Answer
May 14, 2017

Answer:

You need 143 g of #"KClO"_3#.

Explanation:

Step 1. Calculate the moles of #"CO"#

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

#color(blue)(bar(ul(|color(white)(a/a) pV = nRTcolor(white)(a/a)|)))" "#

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 #"O"_2#

The balanced equation is

#"2CO + O"_2 → "2CO"_2#

#"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 #"KClO"_3#

The balanced equation is

#M_text(r):color(white)(ml)122.55#
#color(white)(mmm)"2KClO"_3 → "2KCl" + "3O"_2#

#"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#