If 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amo

Question

If 4.50 grams of ammonia react with 8.80 grams of oxygen what is the maximum amount of nitrogen dioxide that can be made? 12.2 g NO_2 5.03 g NO_2 4.50 g NO_2 7.23 g NO_2 The elements of Group 5A, the nitrogen family, form compounds with hydrogen that have the boiling points listed below. SbH3 - 17 degree C, AsH3 - 55 degree C, PH3 - 87 degree C, NH3 - 33 degree C The first three elements illustrate a trend where the boiling point decreases as the mass decreases; however, ammonia (NH_3) does not follow the trend because of dipole-dipole attraction. London dispersion forces. ionic bonding. hydrogen bonding.

Explanation / Answer

Ans 8 d . 7.23 g NO2

Ammonia reacts with oxygen according to the reaction

4NH3 + 7O2 ---> 4NO2 + 6H2O

4.50 grams of ammonia = 0.264 moles which will require 0.46 moles or 14.8 grams of oxygen.

So oxygen is the limiting reagent.

Now 7 moles or 224 grams of oxygen makes 4 moles or 184 grams of NO2

So 8.80 grams will make 7.23 g of NO2

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