Answer:
Explanation:
<u>1. Molecular chemical equation:</u>
- 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)
<u>2. Mole ratios:</u>
- 2 mol KClO₃ : 2 mol KCl : 3 mol O₂
<u>3. Number of moles of KClO₃</u>
- Number of moles = mass in grams / molar mass
- Molar mass of KClO₃ = 122.55 g/mol
- Number of moles of KClO₃ = 54.3 g / 122.5 g/mol ≈ 0.44308 mol
<u>3. Number of moles of O₂</u>
As per the theoretical mole ratio 2 mol of KClO₃ produce 3 mol of O₂, then set up a proportion to determine how many moles of O₂ will be produced from 0.44038 mol of KClO₃.
- 3 mol O₂ / 2 mol KClO₃ = x / 0.44038 mol KClO₃
- x = (3 / 2) × 0.44308 mol O₂ = 0.6646 mol O₂
Round to 3 significant figures: 0.665 mol of O₂ ← answer
Answer:
HNO2
Explanation:
Because H2CO3 is a weak acid in CH3COOH
Answer:
Theoretical yield of NaCl = 877.5 g or 15 moles
Explanation:
Given data:
Number of moles of Na = 15 mol
Number of moles of Cl₂ = 15 mol
Theoretical yield of NaCl = ?
Solution:
Chemical equation:
2Na + Cl₂ → 2NaCl
Now we will compare the moles of Na and Cl₂ with NaCl.
Cl₂ : NaCl
1 : 2
15 : 2/1×15 = 30 mol
Na : NaCl
2 : 2
15 : 15
Less number of NaCl are produced by Na it will act as limiting reactant.
Theoretical yield:
Mass = number of moles × molar mass
Mass = 15 mol × 58.5 g/mol
Mass = 877.5 g
Answer:
31.9178 °C is the final temperature of the water
Explanation:
Mass of benzene burned = 6.200 g
Moles of benzene burned =
According to reaction , 2 moles of benzene gives 6542 kJ of energy on combustion.
Then 0.0794 mol of benzene on combustion will give:
Mass of water in which Q heat is added = m = 5691 g
Initial temperature =
Final temperature =
Specific heat of water = c = 4.18 J/g°C
Change in temperature of water =
31.9178 °C is the final temperature of the water
False because decaying process should occur in the absence of oxygen