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)
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.
What do you mean convery each of the mols to HNO3 produced?
My NO=.016
My O2=.013
My H20=.66
b. To determine the most Nitric Acid that will be produced in grams, we need to use stoichiometry and the given amounts of reactants. First, we need to convert the masses of nitrogen monoxide (NO) and diatomic oxygen gas (O2) into moles.
1. Calculate the moles of nitrogen monoxide (NO):
Given: mass of NO = 0.5 grams
Molar mass of NO = 30.0 g/mol
moles of NO = mass of NO / molar mass of NO = 0.5 g / 30.0 g/mol
2. Calculate the moles of diatomic oxygen gas (O2):
Given: mass of O2 = 0.4 grams
Molar mass of O2 = 32.0 g/mol
moles of O2 = mass of O2 / molar mass of O2 = 0.4 g / 32.0 g/mol
3. Compare the moles of NO and O2 to determine the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be produced.
The balanced equation from part (a) tells us that the molar ratio between NO and O2 is 4:3. In other words, for every 4 moles of NO, we need 3 moles of O2.
Therefore, we need to find the ratio of moles of NO to moles of O2:
( moles of NO ) / ( moles of O2 ) = ( 0.5 g / 30.0 g/mol ) / ( 0.4 g / 32.0 g/mol )
Simplify the equation:
( moles of NO ) / ( moles of O2 ) = ( 0.5 g * 32.0 g/mol ) / ( 0.4 g * 30.0 g/mol )
( moles of NO ) / ( moles of O2 ) = 16 / 12 = 4 / 3
Since the ratio is 4/3, it means we have equal amounts of moles of NO and O2, so O2 is the limiting reactant.
4. Use the moles of O2 to calculate the moles of Nitric Acid (HNO3) produced:
From the balanced equation, we know that the ratio of O2 to HNO3 is 3:4.
moles of HNO3 = moles of O2 * (4 moles of HNO3 / 3 moles of O2)
Calculating the moles of HNO3:
moles of HNO3 = (0.4 g / 32.0 g/mol) * (4 mol HNO3 / 3 mol O2)
Finally, convert the moles of HNO3 to grams using the molar mass of HNO3 (63.0 g/mol) to get the mass:
mass of HNO3 = moles of HNO3 * molar mass of HNO3
c. To determine the pH of the resulting solution after adding the aqueous Nitric acid to 282 ml of water, we need to know the concentration of nitric acid.
The concentration is typically expressed in moles per liter (molarity, M). In this case, we can assume that the volume remains constant after adding the Nitric acid.
1. Calculate the moles of nitric acid (HNO3) from part b:
Using the moles of HNO3 calculated in part b, divide it by the total volume of the solution (nitric acid + water) to get the concentration in moles per liter (M).
moles of HNO3 = calculated in part b
Total volume of solution = 12 ml (nitric acid) + 282 ml (water)
Convert the total volume to liters:
Total volume = 12 ml + 282 ml = 294 ml = 0.294 L
Concentration (M) = moles of HNO3 / Total volume
2. Now, we have the molarity of nitric acid. We can use this to calculate the pH of the solution. Nitric acid (HNO3) is a strong acid that completely dissociates in water, so the concentration of H+ ions is equal to the concentration of HNO3.
pH = -log10 [H+]
Substitute the concentration of H+ into the equation and calculate the pH.
Note: Ideally, we would need the pKa or pKw values to determine the pH given that other substances like water would also alter the pH of the solution. However, for simplicity, I have assumed that the pH is solely determined by the H+ ions from the nitric acid.