The balanced equation:
2 H₂ + O₂ → 2 H₂O
moles of Hydrogen = mass of hydrogen ÷ molar mass of hydrogen
= 45.7 g ÷ 2 g/mol
= 22.85 mol
Now, the mole ratio of O₂ : H₂ is 1 : 2
which means that there are two times as much moles of H₂ as O₂ taking
part in the reaction
Thus since moles of H₂ = 22.85 mol
then moles of O₂ = 22.85 ÷ 2 = 11.425 mol
Mass of O₂ = moles of O₂ × molar mass of O₂
= 11.425 mol × 32 g/mol
= 365.6 g
∴ <span>grams of oxygen are needed to react with 45.7 grams of hydrogen is 365.6g</span>
Answer:
B. Fluorine
Explanation:
Nonmetallic character increases from <em>bottom to top</em> in a Group and from <em>left to right</em> in a period (see image).
Thus, the most nonmetallic elements are at the <em>upper right corner</em> of the Periodic Table.
We ignore the noble gases, because they are quite unreactive.
So, the element that most “wants” an electron is fluorine.
N, our atmosphere is about 80% nitrogen(N) and about 20% oxygen(O)
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
