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How many grams of O2 are needed to react with 14.8 g of NH3

User CesarB
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Answer:

The balanced equation for the reaction of NH3 and O2 is: 4 NH3 + 5 O2 -> 4 NO + 6 H2O.

From the balanced equation, it can be determined that for every 4 moles of NH3, 5 moles of O2 are needed to react.

To convert 14.8 g of NH3 to moles, the following calculation can be used: 14.8 g NH3 / 17.03 g/mol = 0.87 moles NH3.

Therefore, the number of moles of O2 needed to react with 14.8 g of NH3 is: 0.87 moles NH3 * (5 moles O2 / 4 moles NH3) = 1.09 moles O2.

To convert moles of a substance to grams, the following calculation can be used: moles x molar mass (in g/mol)

Therefore, the number of grams of O2 needed to react with 14.8 g of NH3 is: 1.09 moles O2 * (32.00 g/mol) = 35.1 g O2

User Greatwitenorth
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