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
