Answer:
312 g of O₂
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Next, we shall determine the number of mole of O₂ produced by the reaction of 6.5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 mole of KClO₃ decomposed to 3 moles of O₂.
Therefore, 6.5 moles of KClO₃ will decompose to produce = (6.5 × 3)/2 = 9.75 moles of O₂.
Finally, we shall determine the mass of 9.75 moles of O₂. This can be obtained as follow:
Mole of O₂ = 9.75 moles
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ =?
Mole = mass / Molar mass
9.75 = Mass of O₂ / 32
Cross multiply
Mass of O₂ = 9.75 × 32
Mass of O₂ = 312 g
Thus, 312 g of O₂ were obtained from the reaction.
Answer:
3.1 × 10² g
Explanation:
Step 1: Write the balanced decomposition reaction
KClO₃ ⇒ KCl + 1.5 O₂
Step 2: Calculate the moles of O₂ formed from 6.5 moles of KClO₃
The molar ratio of KClO₃ to O₂ is 1:1.5.
6.5 mol KClO₃ × 1.5 mol O₂/1 mol KClO₃ = 9.8 mol O₂
Step 3: Calculate the mass corresponding to 9.8 moles of O₂
The molar mass of O₂ is 32.00 g/mol.
9.8 mol × 32.00 g/mol = 3.1 × 10² g
