Question

A sample of gas contains 0.1100 mol of H2(g) and 5.500E-2 mol of O2(g) and occupies a volume of 6.69 L. The following reaction takes place:. calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant. 2H2(g) + O2(g)Arrow.gif2H2O(g)
13. A) What volume of methane (CH4) is required to react with 181 liters of carbon tetrachloride according to the following reaction? Answer should be in liters methane (CH4) (All gases are at the same temperature and pressure.) methane (CH4) (g) + carbon tetrachloride (g) Arrow.gifdichloromethane (CH2Cl2) (g)

b) What is the total volume of gaseous products formed when 53.3 liters of ammonia react completely according to the following reaction? Answer is in liters products(All gases are at the same temperature and pressure.) ammonia (g) + oxygen(g)Arrow.gifnitrogen monoxide (g) + water(g)

Answer 1

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 0,1100 moles of H₂O(g)

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

V₂ = 4,46L

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

V₂ = 133L

I hope it helps!