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
The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g
We'll begin by writing the balanced equation for the reaction. This is given below:
2Fe₂O₃ -> 4Fe + 3O₂
- Molar mass of Fe₂O₃ = 159.7 g/mol
- Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
- Molar mass of Fe = 55.85 g/mol
- Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
<h3>How to determine the mass of iron, Fe produced</h3>
From the balanced equation above,
319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe
Therefore,
21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe
Thus, 15.04 g of Fe were produced.
Learn more about stoichiometry:
brainly.com/question/9526265
#SPJ1
Answer : The correct option is, (B) 0.11 M
Solution :
First we have to calculate the concentration 
and 
.
The given equilibrium reaction is,
Initially 0.70 0.70 0
At equilibrium (0.70-x) (0.70-x) x
The expression of 
will be,
![K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BPCl_3%5D%5BCl_2%5D%7D)
Now put all the given values in the above expression, we get:
By solving the term x, we get
From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.
Thus, the concentration of at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of 
at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M
The concentration of 
at equilibrium = x = 0.59 M
Therefore, the concentration of at equilibrium is 0.11 M
To get the ∆S of the reaction, we simply have to add the ∆S of the reactants and the ∆S of the products. Then, we get the difference between the ∆S of the products and the ∆S of the products. If the <span>∆S is negative, then the reaction spontaneous. If the otherwise, the reaction is not spontaneous.</span>
I uploaded the answer to a file hosting. Here's link:
bit.
ly/3a8Nt8n
