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If 50.0 g of KCl reacts with 50.0 g of O2 to produce KClO3 according to the following equation, how many grams of KClO3 will be formed? Word and Formula equation.

If 500 g of KCl reacts with 500 g of O2 to produce KClO3 according to the following equation how many grams of KClO3 will be formed Word and Formula equation class=

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Eduard22sly

Answer:

A. 82.2g of KClO3

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3

C. Formula equation:

2KCl + 3O2 —> 2KClO3

Explanation:

The balanced equation for the reaction. This is given below:

2KCl + 3O2 —> 2KClO3

Next, we shall determine the masses of KCl and O2 that reacted and the mass of KClO3 produced from the balanced equation. This is illustrated below:

Molar Mass of KCl = 39 + 35.5 = 74.5g/mol

Mass of KCl from the balanced equation = 2 x 74.5 = 149g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96g

Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 122.5g/mol

Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g

Summary:

From the balanced equation above:

149g of KCl reacted.

96g of O2 reacted.

245g of KCl were produced.

Next, we shall determine the limiting reactant. This is illustrated below:

From the balanced equation above,

149g of KCl reacted with 96g of O2.

Therefore, 50g of KCl will react with = (50 x 96)/149 = 32.21g of O2.

Since a lesser mass of O2 ( i.e 32.21g) than what was given (i.e 50g) is needed to react completely with 50g of KCl, therefore, KCl is the limiting reactant and O2 is the excess reactant.

A. Determination of the mass of KClO3 produced from the reaction.

In this case the limiting reactant will be used.

From the balanced equation above,

149g of KCl reacted To produce 245g of KClO3.

Therefore, 50g of KCl will react to produce = (50 x 245)/149 = 82.2g of KClO3.

Therefore, 82.2g of KClO3 is produced from the reaction.

B. Word equation:

50g of KCl react with 50g of O2 to produce 82.2g of KClO3.

C. Formula equation:

2KCl + 3O2 —> 2KClO3

The mass of [tex]\rm KClO_3[/tex] produced has been 82.1085 g.

The word equation for the reaction has been: 50 gram potassium chloride reacts with 50 g oxygen to give 82.1085 g potassium chlorate.

The formula equation has been, [tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex].

The balanced chemical equation for the reaction has been given as:

[tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex]

Computation for Mass of Potassium chlorate:

From the balanced chemical equation, 2 moles of KCl reacts with 3 moles of oxygen to give 2 moles of [tex]\rm KClO_3[/tex].

The moles of reactants have been given as:

[tex]\rm Moles=\dfrac{mass}{molar\;mass}[/tex]

The moles of 50 g KCl has been given as:

[tex]\rm Moles\;KCl=\dfrac{50}{74.55} \\Moles\;KCl=0.67\;mol[/tex]

The moles of KCl available has been 0.67 mol.

The moles of 50 g [tex]\rm O_2[/tex] has been given as:

[tex]\rm Moles\;O_2=\dfrac{50}{32}\\Moles\;O_2=1.5625\;mol[/tex]

The moles of 50 g [tex]\rm O_2[/tex] has been 1.5625 mol.

From the balanced equation, for 2 moles KCl, 3 moles oxygen has been required. For 0.67 mol KCl, oxygen required has been:

[tex]\rm 2\;mole\;KCl=3\;moles\;O_2\\0.67\;mol\;KCl=\dfrac{3}{2}\;\times\;0.67\;mol\;O_2\\0.67\;mol\;KCl=1.005\;mol\;O_2[/tex]

The available moles of oxygen has been 1.5625 mol. Thus, oxygen has been excess reactant and KCl has been limiting reactant.

The moles of [tex]\rm KClO_3[/tex] produced has been given as:

[tex]\rm 2\;mol\;KCl=2\;mol\;KClO_3\\0.67\;mol\;KCl=\dfrac{2}{2}\;\times\;0.67\;mol\;KClO_3\\0.67\;mol\;KCl=0.67\;mol\;KClO_3[/tex]

The moles of [tex]\rm KClO_3[/tex] produced has been 0.67 mol.

The mass of [tex]\rm KClO_3[/tex] produced has been:

[tex]\rm Mass=Moles\;\times\;molar\;mass\\Mass\;KCl=0.67\;\times\;122.55\;\\Mass\;of\;KCl=82.1085\;g[/tex]

The mass of [tex]\rm KClO_3[/tex] produced has been 82.1085 g.

The word equation for the reaction has been:

50 gram potassium chloride reacts with 50 g oxygen to give 82.1085 g potassium chlorate.

The formula equation has been, [tex]\rm 2\;KCl\;+\;3\;O_2\;\rightarrow\;2\;KClO_3[/tex].

For more information about chemical equation, refer to the link:

https://brainly.com/question/13350862

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