If 7.45g of N2 and 36.5mL of 0.150M O2 react together, how many grams of N2O gas will be produced?

Question

If 7.45g of N2 and 36.5mL of 0.150M O2 react together, how many grams of N2O gas will be produced?

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Amity 1 year 2021-09-03T09:24:38+00:00 1 Answers 10 views 0

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    2021-09-03T09:25:59+00:00

    Answer:

    0.482 g of N₂O are produced.

    Explanation:

    The reaction is: 2N₂ + O₂ →  2N₂O

    In order to produce 2 moles of N₂O, we need to make react 1 mol of oxygen and 2 moles of nitrogen.

    We determine the moles of each reactant:

    7.45 g / 28g/mol = 0.266 moles of nitrogen

    36.5 mL . 0.15 M = 5.475 mmoles →  . 1mol /1000 mmol = 0.00547 moles

    Certainly the oxygen is the limiting reactant.

    In order to determine the grams produced, we need to know the limiting reactant.

    2 moles of N₂ need 1 mol of O₂

    Then 0.266 moles of N₂ may react to (0.266 . 1) /2 = 0.133 moles

    We only have 0.00547 moles, so there is not enough O₂.

    1 mol of O₂ can produce 2 moles of N₂O

    Then, the 0.00547 will produce (0.00547 . 2) /1 = 0.01095 moles

    We convert to mass: 0.01095 mol . 44 g /mol = 0.482 g

    That’s the answer.

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