For the following reaction, 5.04 grams of nitrogen gas are allowed to react with 8.98 grams of oxygen gas: nitrogen(g) + oxygen(g) --> nitrogen monoxide(g)
a.What is the maximum mass of nitrogen monoxide that can be formed?
b. What is the FORMULA for the limiting reagent?
c. What mass of the excess reagent remains after the reaction is complete
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Ответ:
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.
Ответ: