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
Electronegativity is your answer.
Answer:
0.712 mol of NO₂ are formed .
Explanation:
For the reaction , given in the question ,
2 N₂O₅ ( g ) → 4 NO₂ ( g ) + O₂ ( g )
From the above balanced reaction ,
2 mol of N₂O₅ reacts to give 4 mol of NO₂
Applying unitary method ,
1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂
From the question , 0.356 mol of N₂O₅ are reacted ,
<u>now, using the above equation , to calculate the moles of the NO₂ , as follow -</u>
Since ,
1 mol of N₂O₅ reacts to give 4 / 2 mol of NO₂
0.356 mol of N₂O₅ reacts to give 4 / 2 * 0.356 mol of NO₂
Calculating ,
0.712 mol of NO₂ are formed .
