a. The equation looks ok to me.
b.
mols NO = 0.5g/molar mass NO = ?
mols O2 = 0.4 g/molar mass NO = ?
mols H2O = 12/molar mass H2O = ?
Now convert each of the mols above to mols HNO3 produced. This is a limiting reagent (LR) problem and the smaller number of mols will be produced and the reagent responsible for that is the LR. The smaller(or smallest number) x molar mass HNO3 = grams HNO3
c. Actually the volume probably are not additive but the problem tells us they are which makes it easier; However, what is the volume of the HNO3 produced. There is no way of knowing unless you know the density of the solution and the problem doesn't tell you that. Technically you can't solve the part. Assuming the density is 1.0 (and it isn't) then grams HNO3 = volume in mL and that + 282 mL = = total volume in mL. Change to L, then mols/L = M
Finally, pH = -log(HNO3)
Post your wo0rk if you get stuck.
Situation: The Production of Nitric Acid can be achieved by the following process:
Nitrogen monoxide gas(NO) + diatomic oxygen gas+ water (liquid) ===> Nitric Acid (aq)
a. Show complete balanced reaction for process
What I have: 4NO+ 3O2+ 2H2O > 4HNO3
b. The lab technician combines .5 grams of nitrogen monoxide with .4 grams of diatomic oxygen gas and 12ml of water. What is the most Nitric Acid that will be produced in grams?
c. Resulting aqueous Nitric acid is then added to 282 ml of water. What is the pH? (Caution: aqueous volume of nitric acid and 282 ml of new water are additive)
2 answers
What do you mean convery each of the mols to HNO3 produced?
My NO=.016
My O2=.013
My H20=.66
My NO=.016
My O2=.013
My H20=.66
1 answer