Molar mass N₂ = 28.0 g/mol
1 mole ------------- 28.0 g
5 moles ------------ ?
mass = 5 . 28.0 / 1
mass = 140 g
hope this helps!
The balanced equation says that 2 moles C₂H₆ and 7 moles O₂ react together, i.e. in a ratio of 7:2 or 3.5 moles of O₂ to C₂H₆.
With molar masses 30.07 g/mol (C₂H₆) and 31.998 g/mol (O₂), the given quantities amount to
(19 g C₂H₆) × (1/30.07 mol/g) ≈ 0.63 mol C₂H₆
(115 g O₂) × (1/31.998 mol/g) ≈ 3.59 mol O₂
Now, 0.63/2 ≈ 0.32, and for every 0.32 mol C₂H₆ consumed, the reaction requires 7×0.32 ≈ 2.2 mol O₂. Then in order to consume all of the C₂H₆, the reaction would need 2×2.2 ≈ 4.4 mol O₂, which we don't have.
In other words, we have too much C₂H₆ and not enough O₂, so O₂ is the limiting reactant.
Answer:
Organic waste in landfills generates, methane, a potent greenhouse gas. By composting wasted food and other organics, methane emissions are significantly reduced. Compost reduces and in some cases eliminates the need for chemical fertilizers. Compost promotes higher yields of agricultural crops.
Explanation:
Explanation:
The given reaction equation will be as follows.
![[FeSCN^{2+}] \rightleftharpoons [Fe^{3+}] + [SCN^{-}]](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%5Crightleftharpoons%20%5BFe%5E%7B3%2B%7D%5D%20%2B%20%5BSCN%5E%7B-%7D%5D)
Let is assume that at equilibrium the concentrations of given species are as follows.
![[Fe^{3+}] = 8.17 \times 10^{-3}](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%20%3D%208.17%20%5Ctimes%2010%5E%7B-3%7D)
M
![[SCN^{-}] = 8.60 \times 10^{-3}](https://tex.z-dn.net/?f=%5BSCN%5E%7B-%7D%5D%20%3D%208.60%20%5Ctimes%2010%5E%7B-3%7D)
M
![[FeSCN^{2+}] = 6.25 \times 10^{-2}](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%3D%206.25%20%5Ctimes%2010%5E%7B-2%7D)
M
Now, first calculate the value of 
as follows.
![K_{eq} = \frac{[Fe^{3+}][SCN^{-}]}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BFe%5E%7B3%2B%7D%5D%5BSCN%5E%7B-%7D%5D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
= 
= 
Now, according to the concentration values at the re-established equilibrium the value for ![[FeSCN^{2+}]](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D)
will be calculated as follows.
![11.24 \times 10^{-4} = \frac{8.12 \times 10^{-3} \times 7.84 \times 10^{-3}}{[FeSCN^{2+}]}](https://tex.z-dn.net/?f=11.24%20%5Ctimes%2010%5E%7B-4%7D%20%3D%20%5Cfrac%7B8.12%20%5Ctimes%2010%5E%7B-3%7D%20%5Ctimes%207.84%20%5Ctimes%2010%5E%7B-3%7D%7D%7B%5BFeSCN%5E%7B2%2B%7D%5D%7D)
![[FeSCN^{2+}] = 5.66 \times 10^{-2}](https://tex.z-dn.net/?f=%5BFeSCN%5E%7B2%2B%7D%5D%20%3D%205.66%20%5Ctimes%2010%5E%7B-2%7D)
M
Thus, we can conclude that the concentration of in the new equilibrium mixture is 
M.
Answer:
by filtering it with filter paper
