Answer :
Answer:
1. 10.8 g of NO
2. N₂ is the limting reagent
3. 3.2 g of O₂ does not react
Explanation:
We determine the reaction: N₂(g) + O₂(g) → 2NO(g)
We need to determine the limiting reactant, but first we need the moles of each:
5.04 g / 29 g/mol = 0.180 moles N₂
8.98 g / 32 g/mol = 0.280 moles O₂
Ratio is 1:1, so the limiting reactant is the N₂. For 0.280 moles of O₂ I need the same amount, but I only have 0.180 moles of N₂
Ratio is 1:2. 1 mol of N₂ can produce 2 moles of NO
Then, 0.180 moles of N₂ may produce (0.180 .2) / 1 = 0.360 moles NO
If we convert them to mass → 0.360 mol . 30 g/1 mol = 10.8 g
As ratio is 1:1, for 0.180 moles of N₂, I need 0.180 moles of O₂.
As I have 0.280 moles of O₂, (0.280 - 0.180 ) = 0.100 moles does not react.
0.1 moles . 32 g/mol = 3.2 g of O₂ remains after the reaction is complete.
Answer:
(a) 10.8 grams of nitrogen monoxide
(b) The formula for the limiting reagent is N
(c) 3.22 grams of the excess reagent remains
Explanation:
(a) The equation of reaction involves the reaction of 2 moles of nitrogen atom (N) and 1 mole of oxygen molecule (O2) to form 2 moles of nitrogen monoxide (NO).
From the equation of reaction above,
2 moles of nitrogen atom (28 grams) produced 2 moles of nitrogen monoxide (60 grams)
5.04 grams of nitrogen would produce (5.04×60)/28 = 10.8 grams of nitrogen monoxide.
(b) 1 mole of oxygen (32 grams) reacted with 2 moles of nitrogen (28 grams)
8.98 grams of oxygen would be required to react with (28×8.98)/32 = 7.86 grams of nitrogen but only 5.04 grams is available. Therefore, the limiting reagent is nitrogen atom and its formula is N.
(c) 2 moles of nitrogen monoxide (60 g) are produced from 1 mole of oxygen (32 g)
10.8 g of nitrogen monoxide is produced from (10.8×32)/60 = 5.76 grams of oxygen but 8.98 grams of oxygen is available. Therefore, oxygen is the excess reagent.
Mass of excess reagent that remains after the reaction is complete = 8.98 - 5.76 = 3.22 grams