Chemistry, 19.01.2021 06:10 CalCDanG

In an experiment, potassium chlorate decomposed according to the following chemical equation. KClO3 → KCl + O2

(Molar mass of KClO3 = 122.5 g/mol; KCl = 74.55 g/mol; O2 = 31.998 g/mol)

If the mass of potassium chlorate was 240 grams, which of the following calculations can be used to determine the mass of oxygen gas formed?

(240 × 2 × 31.998) ÷ (122.5 × 3) grams
(240 × 3 × 31.998) ÷ (122.5 × 2) grams
(240 × 2 × 122.5) ÷ (31.998 × 3) grams
(240 × 3 × 122.5) ÷ (31.998 × 2) grams

Answers: 3

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Answer from: kdamaso93

In the experiment the potassium and chlorate decompose as per the chemical equation of KClO3 → KCl + O2.

Hence the method to use the mole concept approach. Here find the number of moles of the given specie and use it to find that of the unknown. Then convert the number of moles of the unknown to mass. Mass = number of moles x molar massThus (240 × 3 × 31.998) ÷ (122.5 × 2) grams.

Learn more about the potassium chlorate decomposed according.

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Answer from: kieramacphee6216

(240 × 3 × 31.998) ÷ (122.5 × 2) grams

Explanation:

To solve this problem, it is best to use the mole concept approach. Here find the number of moles of the given specie and use it to find that of the unknown.

Then convert the number of moles of the unknown to mass.

Number of moles = $\frac{mass}{molar mass}$

Mass = number of moles x molar mass

Mass of potassium chlorate = 240g

The balanced reaction equation is given as:

2KClO₃ → 2KCl + 3O₂

Number of moles of KClO₃ = $\frac{240}{122.5}$

Mole ratio:

2 mole of KClO₃ will produce 3 moles of O₂

$\frac{240}{122.5}$ mole of KClO₃ will produce $\frac{240}{122.5}$ x 3 / 2 mole of O₂

So, mass of oxygen;

Mass = $\frac{240}{122.5}$ x 3 / 2 x 31.998 = $\frac{240}{122.5}$ x 3 x $\frac{1}{2}$ x 31.998

This gives (240 × 3 × 31.998) ÷ (122.5 × 2) grams

Answer from: Quest

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is $1s^{2}2s^{2}2p^{6}3s^{1}$ and electronic configuration of phosphorous is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{3}$ .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

$3na + p \rightarrow na_{3}p$

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.