Answer:
- 10.5 g of N₂
- Limiting reagent: NO
- 3.13 g of H₂ remains
Explanation:
First of all we state the reaction: 2NO(g) + 2H₂(g) → 2H₂O(l) + N₂(g)
We need to find out the limiting reactant and the excess reagent
Ratio in the reactants is 2:2. Let's convert the mass to moles:
22.6 g / 30 g/mol = 0.753 moles of NO
4.64 g / 2 g/mol = 2.32 moles of H₂
Certainly the limiting reagent is the NO and the excess reactant is the hydrogen:
- For 0.753 moles of NO, we need 0.753 moles of H₂ (we have 2.32 moles)
- For 2.32 moles of H₂, we need 2.32 moles of NO (and we don't have enough NO, because we only have 0.753 moles)
As the H₂ is the excess reagent, some moles still remains after the reaction is complete → 2.32 mol - 0.753 mol = 1.567 moles
We convert the moles to mass: 1.567 mol . 2g /1mol = 3.13 g of H₂ remains
As the NO is the limiting reagent, we can work with the equation:
We propose this rule of three: 2 moles of NO can produce 1 mol of N₂
Then, 0.753 moles of NO must produce (0.753 . 1) /2 = 0.376 moles of N₂
We convert the moles to mass 0.376 mol . 28 g / 1 mol = 10.5 g
