A sample of $KClO_3$ on decomposition gives 448 $mL$ of oxygen gas at NTP. Calculate: (a) Weight of oxygen produced (b) Weight of $KClO_3$ originally taken (c) Weight of $KCl$ produced?

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A sample of $KClO_3$ on decomposition gives 448 $mL$ of oxygen gas at NTP. Calculate: (a) Weight of oxygen produced (b) Weight of $KClO_3$ originally taken (c) Weight of $KCl$ produced?

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Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

Start with the balanced chemical equation for the of potassium chlorate, $KClO_3$

$color(green)(2)KClO_(3(s)) stackrel(color(red)("heat"))(->) color(blue)(2)KCl_((s)) + color(orange)(3)O_(2(g))$

The important thing to notice here is that $color(green)(2)$ moles of potassium chlorate produce $color(blue)(2)$ moles of potassium chloride and $color(orange)(3)$ moles of oxygen gas.

Since you know the volume and conditions under which the oxygen gas was produced, you can use the equation to determine how many moles were produced.

At NTP, or Normal Temperature and Pressure, the pressure is set at 1 atm and the temperature at $20^@"C"$.

$PV = nRT => n = (PV)/(RT)$

$n_(O_2) = (1cancel("atm") * 448 * 10^(-3)cancel("L"))/(0.082(cancel("atm") * cancel("L"))/("mol" * cancel("K")) * (273.15 + 20)cancel("K")) = "0.01864 moles"$ $O_2$

Use oxygen's molar mass to determine how many grams were produced

$0.01864cancel("moles") * "32.0 g"/(1cancel("mole")) = color(green)("0.596 g " O_2)$

SIDE NOTE Do not forget to convert the temperature to Kelvin and the volume to liters!

The mass of potassium chlorate that underwent decomposition can be found in a similar manner. First, determine the number of moles by using ratio that exists between potassium chlorate and oxygen gas

$0.01864cancel("moles"O_2) * (color(green)(2)" moles "KClO_3)/(color(orange)(3)" moles "O_2) = "0.01243 moles"$ $KClO_3$

The mass of the compound before decomposition was

$0.01243cancel("moles") * "122.55 g"/(1cancel("mole")) = color(green)("1.52 g "KClO_3)$

The number of moles of potassium chlorate produced by the reaction will be equal to the number of moles of potassium chlorate

$0.01243cancel("moles"KClO_3) * (color(blue)(2)" moles "KCl)/(color(green)(2)" moles "KClO_3) = "0.01243 moles"$ $KCl$

The mass of potassium chloride is

$0.01243cancel("moles") * "74.55 g"/(1cancel("mole")) = color(green)("0.927 g "KCl)$

All three values are rounded to three , the number of sig figs you gave for the volume of oxygen gas produced.

A sample of $KClO_3$ on decomposition gives 448 $mL$ of oxygen gas at NTP. Calculate: (a) Weight of oxygen produced (b) Weight of $KClO_3$ originally taken (c) Weight of $KCl$ produced?

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