Given :
Number of moles , n = 36.25 mol .
Molarity , M = 1.25 M .
To Find :
The volume of water required .
Solution :
Moarity is given by :
So ,
Here , n is number of moles and M is molarity .
Putting all values in above equation , we get :
Therefore , volume of water required is 29 L .
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