When CH₄ is burnt in excess O₂ following products are formed,
CH₄ + 2 O₂ → CO₂ + 2 H₂O
According to equation 1 mole of CH₄ (16 g) reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O. Hence the products are,
1 mole of CO₂ and 2 moles of H₂O
Converting 1 mole CO₂ to grams;
As,
Mass = Moles × M.mass
Mass = 1 mol × 44 g.mol⁻¹
Mass = 40 g of CO₂
Converting 2 moles of H₂O to grams,
Mass = 2 mol × 18 g.mol⁻¹
Mass = 36 g of H₂O
Total grams of products;
Mass of CO₂ = 44 g
+ Mass of H₂O = 36 g
-------------
Total = 80 g of Product
Result:
80 grams of product is formed when 16 grams of CH₄ is burnt in excess of Oxygen.
Answer:
0.44 moles
Explanation:
Given that :
A mixture of water and graphite is heated to 600 K in a 1 L container. When the system comes to equilibrium it contains 0.17 mol of H2, 0.17 mol of CO, 0.74 mol of H2O, and some graphite.
The equilibrium constant ![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
The equilibrium constant 
The equilibrium constant 
Some O2 is added to the system and a spark is applied so that the H2 reacts completely with the O2.
The equation for the reaction is :

Total mole of water now = 0.74+0.17
Total mole of water now = 0.91 moles
Again:
![K_c= \dfrac{[CO][H_2]}{[H_2O]}](https://tex.z-dn.net/?f=K_c%3D%20%20%5Cdfrac%7B%5BCO%5D%5BH_2%5D%7D%7B%5BH_2O%5D%7D)
![0.03905 = \dfrac{[0.17+x][x]}{[0.91 -x]}](https://tex.z-dn.net/?f=0.03905%20%3D%20%20%5Cdfrac%7B%5B0.17%2Bx%5D%5Bx%5D%7D%7B%5B0.91%20-x%5D%7D)
0.03905(0.91 -x) = (0.17 +x)(x)
0.0355355 - 0.03905x = 0.17x + x²
0.0355355 +0.13095
x -x²
x² - 0.13095
x - 0.0355355 = 0
By using quadratic formula
x = 0.265 or x = -0.134
Going by the value with the positive integer; x = 0.265 moles
Total moles of CO in the flask when the system returns to equilibrium is :
= 0.17 + x
= 0.17 + 0.265
= 0.435 moles
=0.44 moles (to two significant figures)
24. F:H :) because Fluorine is the most electronegative
A) cesium chloride
B) barium oxide
C) potassium sulfide
D) beryllium chloride
E) hydrogen bromide
F) aluminum fluoride