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isabelvaldez123
25.03.2020 •
Chemistry
For the following reaction, 22.6 grams of nitrogen monoxide are allowed to react with 4.64 grams of hydrogen gas . nitrogen monoxide(g) + hydrogen(g) nitrogen(g) + water(l) What is the maximum amount of nitrogen gas that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit AnswerRetry Entire Group
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Ответ:
- 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
Ответ:
At least that's on the one I have.
I hope that this helps :-)