Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
Answer:
1)
Explanation:
the answer to you question Is 1)
Answer:
Initially
of nitrogen dioxide were in the container .
Explanation:
Volume of the container at low pressure and at room temperature =
Number of moles in the container = 
After more addition of nitrogen gas at the same pressure and temperature.
Volume of the container after addition = 
Number of moles in the container after addition=
Applying Avogadro's law:
(at constant pressure and temperature)



Initially
of nitrogen dioxide were in the container .
Answer:
81.5 L
Explanation:
We can use the combined gas law equation that gives the relationship among pressure, temperature and volume of gases for a fixed amount of gas.
P1V1 / T1 = P2V2 / T2
where P1 - pressure, V1 - volume and T1 - temperature at the first instance
P2 - pressure, V2 - volume and T2 - temperature at the second instance
substituting the values in the equation
1240 Torr x 47.2 L / 298 K = 730 Torr x V2 / 303 K
V2 = 81.5 L
the new volume the gas would occupy when the conditions have changed is 81.5 L