4NH3+5O2=4NO+6H2O how many moles of oxygen gas would be required to produce 135.1 grams of nitrogen monoxide

Answer 1

The molar ratio of oxygen gas to nitrogen monoxide in the balanced equation is 5 moles O2 / 4 moles NO.

As we know the grams of NO required, we need to convert grams NO to moles NO then multiply the answer by the molar ratio 5/4.

Moles NO = grams NO / mol. mass of NO = 135.1 g / 30.01 g/mole = 4.502 moles NO produced.

5 moles O2 / 4 moles NO x 4.502 moles NO = 5.628 moles of O2 required.