O3 + NO --> O2 + NO2 If 0.740 g of O3 reacts with 0.670 g of NO, how many grams of NO2 will be produced?

Question

which compound is the limiting reagent? calculate the number of moles of the excess reagent remaining at the end of the reaction.

Answer 1

Moles O3 = 0.740 g / 48 g/mol =  0.0154

Moles NO = 0.670 g / 30 g/mol =  0.0223

the ratio is 1 : 1 so O3 is the limiting reactant

Moles NO in excess = 0.0223 - 0.0154 = 0.00690