Answer:
4.46L
A. 181L
B. 133L
Explanation:
For the reaction:
2H₂(g) + O₂(g) → 2H₂O(g)
0,1100 moles of H₂ with 0,0055 moles of O₂ produce <em><u>0,1100 moles of H₂O(g)
</u></em>
Using Avogadro's law that says that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. It is possible to write:
V₁ / n₁ = V₂ / n₂
Where:
V₁ is initial volume (6.69L); / n₁ are initial moles (0.1100moles+0.00550moles); V₂ is final volume and n₂ final moles (0.1100moles)
6.69L / 0.165moles = V₂ / 0.1100moles
<em>V₂ = 4,46L</em>
A. For the reaction:
CH₄(g) + CCl₄(g) → 2CH₂Cl₂(g)
The volume of methane you require is 181L. Again, you can use Avogadro's law to know that equal volumes have the same number of molecules. As the reaction is 1:1, volume of methane must be the same of carbon tetrachloride.
B. For the reaction:
4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
You can use, again, Avogadro's law. If 4 moles of reaction that occupy 53,3L produce 10 moles of gases that occupy...:
V₁ / n₁ = V₂ / n₂
Where:
V₁ is initial volume (53,3L); / n₁ are initial moles (4moles); V₂ is final volume and n₂ final moles (10)
53.3L / 4moles = V₂ / 10moles
<em>V₂ = 133L</em>
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I hope it helps!
