How many grams of NO will be produced from 75.1 g of NO₂ reacted with excess water in the following chemical reaction? 3 NO₂(g) + H₂O(l) → 2 HNO₃(g) + NO(g)

Answer 1

This reaction is a simple stoichiometric conversion. We know that one of the reactants, water, is excess, so NO₂ will be the limiting reagent in this reaction. First we will convert the grams of NO₂ into moles by diving by its molar mass as follows.

75.1g NO₂/ 46.01g/mol NO₂=1.63 moles NO₂

Then we will use the mole ratio that 3 moles of NO₂= 1 mole NO to convert moles of NO₂ to moles of NO.

1.63 moles NO₂ /(1mol NO / 3mol NO₂ )=.544 moles NO

Finally, we will convert moles of NO to grams of NO by multiplying the amount of moles by the molar mass of NO as follows.

.544mols NO* 30.01g/mol NO= 16.3g NO