What volume of carbon dioxide gas can be collected from
1 answer:
Answer:
1.22 L of carbon dioxide gas
Explanation:
The reaction that takes place is:
- CaCO₃ + HCl → CaCl₂ + CO₂ + H₂O
First we <u>determine which reactant is limiting</u>:
- Calcium carbonate ⇒ 10.0 g CaCO₃ ÷ 100 g/mol = 0.10 mol CaCO₃
- Hydrochloric acid ⇒ 0.100 L * 0.50 M = 0.05 mol HCl
So HCl is the limiting reactant.
Now we calculate the moles of CO₂ produced:
- 0.05 mol HCl *
= 0.05 mol CO₂
Finally we use PV=nRT to <u>calculate the volume</u>:
- P = 1 atm
- n = 0.05 mol
- T = 25 °C ⇒ 25 + 273.16 = 298.16 K
1 atm * V = 0.05 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K
- V = 1.22 L
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Answer:
The mass of oxygen is 12.10 g.
Explanation:
The decomposition reaction of potassium chlorate is the following:
2KClO₃(s) → 2KCl(s) + 3O₂(g)
We need to find the number of moles of KClO₃:
Where:
m: is the mass = 30.86 g
M: is the molar mass = 122.55 g/mol
Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3
Finally, the mass of O₂ is:
Therefore, the mass of oxygen is 12.10 g.
I hope it helps you!
Answer:
300.9mL
Explanation:
Given parameters:
V₁ = 280mL
T₁ = 22°C
T₂ = 44°C
Unknown:
V₂ = ?
Solution:
To solve this problem, we apply Charles's law;
it is mathematically expressed as;
We need to convert the temperature to kelvin;
T₁ = 22°C = 22 + 273 = 295K
T₂ = 44°C = 44 + 273 = 317K
Input the parameters and solve;
=
V₂ x 295 = 280 x 317
V₂ = 300.9mL