Answer:
the answer is 25.032g check pic for explanation
Answer:
The amount of NO₂ that can be produced 8.533 g
Explanation:
According to question
2 NO(g) + O₂(g) → 2 NO₂(g)
Given
Moles of nitrogen monoxide = 0.377
Moles of oxygen = 0.278
Since 'NO' is the limiting reagent according to this ratio.
According to equation
2 moles NO reacts to form 2 moles NO₂
So, 0.1855 moles NO give = 0.1855 moles of NO₂
Mass of 1 mole NO₂ = 46 g/mole
Mass of 0.1855 moles = 46 x 0.1855 = 8.533 g
Answer: [N2]₀ = 10M and [H2]₀ = 11M
Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:
I is initial amount;
C is change in concentration;
E is for equilibrium concentration;
For the mixture,
N2 3H2 2NH3
I [N2]₀ [H2]₀ 0
C - x -3x +2x
E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4
With the product, we can find "x":
2x=4
x=2M
With x=2, find the concentrations:
[N2]₀ - x = 8
[N2]₀ = 10M
[H2]₀ - 3x = 5
[H2]₀ = 11M
The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.