calculate the mass of sodium trioxonitrate(v) produced when 3.0g of pure sodium hydroxide reacts with 100cm3 of 1.0m trioxonitrate(v) acid what is the solution
NaOH + HNO3 = NaNO3 + H2O
100cm^3 of 1M HNO3 contains 0.1 mole of solute
3.0g of NaOH is 3/40 = 0.075 moles
so, you will get 0.075 moles of NaNO3
I do not believe trioxonitrate(v) acid is a correct IUPAC name for nitric acid (HNO3). That name has become relatively common on this site in about the last five years. A correct IUPAC name is nitric acid for the formula HNO3. As best I can determine, if you insist on the screwball system, then it should be called trioxidonitrate (not trioxo) and that is acceptable as an IUPAC name also. For the life of me, however, I just can't understand why we try to complicate matters so much. Chemistry is not one of those household careers. It is hard enough as is; why do we need to make it more intimidating. I have heard hundreds of students who don't want to take chemistry because it's "too hard". So we make it more complicated instead of less so. Too bad!
To calculate the mass of sodium trioxonitrate(V) produced, we need to determine the limiting reactant in the given reaction. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
First, let's find the number of moles of sodium hydroxide (NaOH) using its molar mass. The molar mass of NaOH is 22.99 g/mol for sodium (Na) + 16.00 g/mol for oxygen (O) + 1.01 g/mol for hydrogen (H), which gives a total of 39.99 g/mol for NaOH.
Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
= 3.0 g / 39.99 g/mol
= 0.075 moles
Next, we need to find the number of moles of trioxonitrate(V) acid (HNO3). Since the concentration of the acid is given as 1.0 M (1.0 mol/L), we can directly calculate the number of moles using the volume.
Number of moles of HNO3 = Concentration of HNO3 x Volume of HNO3
= 1.0 mol/L x 0.1 L (since 100 cm3 is equal to 0.1 L)
= 0.1 moles
The balanced chemical equation for the reaction between NaOH and HNO3 is:
NaOH + HNO3 -> NaNO3 + H2O
From the balanced equation, we can see that the molar ratio between NaOH and NaNO3 is 1:1. Therefore, the number of moles of NaNO3 produced is also 0.075 moles.
Finally, we calculate the mass of NaNO3 using its molar mass. The molar mass of NaNO3 is 22.99 g/mol for sodium (Na) + 14.01 g/mol for nitrogen (N) + 48.00 g/mol for oxygen (O), which gives a total of 85.00 g/mol for NaNO3.
Mass of NaNO3 = Number of moles of NaNO3 x Molar mass of NaNO3
= 0.075 moles x 85.00 g/mol
= 6.37 g
Therefore, the mass of sodium trioxonitrate(V) produced when 3.0 g of pure sodium hydroxide reacts with 100 cm3 of 1.0 M trioxonitrate(V) acid is approximately 6.37 g.
To calculate the mass of sodium trioxonitrate(V) produced, we first need to write out the balanced chemical equation for the reaction:
2NaOH + HNO3 -> NaNO3 + 2H2O
From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of HNO3 to produce 1 mole of NaNO3.
Step 1: Convert grams of NaOH to moles.
The molar mass of NaOH is 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol.
To convert grams to moles, we will use the formula:
moles = mass / molar mass
moles of NaOH = 3.0 g / 39.00 g/mol ≈ 0.077 moles
Step 2: Convert volume of HNO3 to moles.
The molarity (M) of HNO3 solution is given as 1.0 M, which means 1 mole of HNO3 is present in 1 liter of the solution.
To convert 100 cm3 of HNO3 solution to liters, we will use the conversion factor:
1 L = 1000 cm3
volume of HNO3 = 100 cm3 / 1000 cm3/L = 0.1 L
moles of HNO3 = molarity x volume in liters
moles of HNO3 = 1.0 M x 0.1 L = 0.1 moles
Step 3: Determine the limiting reactant.
The limiting reactant is the one that is completely consumed and determines the amount of product formed. To determine the limiting reactant, we compare the moles of each reactant.
From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of HNO3. Therefore, the ratio of moles of NaOH to moles of HNO3 is 2:1.
Since we have 0.077 moles of NaOH and 0.1 moles of HNO3, it means that we have excess HNO3 and NaOH is the limiting reactant.
Step 4: Calculate the moles of NaNO3 produced.
From the balanced equation, we know that 2 moles of NaOH react to give 1 mole of NaNO3.
moles of NaNO3 produced = (moles of NaOH) / 2 = 0.077 moles / 2 = 0.0385 moles
Step 5: Convert moles of NaNO3 to grams.
The molar mass of NaNO3 is 22.99 g/mol + 14.01 g/mol + 48.00 g/mol = 85.00 g/mol.
mass of NaNO3 = moles of NaNO3 x molar mass
mass of NaNO3 = 0.0385 moles x 85.00 g/mol ≈ 3.29 grams
Therefore, the mass of sodium trioxonitrate(V) produced is approximately 3.29 grams.