What volume (in mL) of a 0.150 M HNO3 solution will completely react with 35.7 mL of a 0.108 M Na2CO3 solution according to the following balanced chemical equation?
Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
A. Express the volume in milliliters to three significant figures.
the volume of HNO3 solution = 51.4mL
B. In the reaction in Part A, what mass (in grams) of carbon dioxide forms?
Na2CO3(aq)+2HNO3(aq)→2NaNO3(aq)+CO2(g)+H2O(l)
millimoles Na2CO3 = mL x M = 35.7 x 0.108 = 3.856
millimoles HNO3 needed = 3.855 x (2 mol HNO3/1 mol Na2CO3) = 7.711
7.711 mmoles HNO3 = mL x M so
mL = 7.711/0.150 = 51.408 = 51.4 mL
millimoles CO2 = 3.856 mmol Na2CO3 x (1 mol CO2/1 mol Na2CO3) = 3.856 or 0.003856 mols, then grams CO2 = mols CO2 x molar mass CO2 = ? and remember to round to 3 s.f.
A. Well, let's calculate the volume of the HNO3 solution, shall we? According to the balanced equation, we have a 1:2 ratio between Na2CO3 and HNO3. So, if 35.7 mL of 0.108 M Na2CO3 solution reacts, we need twice that amount of HNO3. That means 35.7 mL * 2 = 71.4 mL of HNO3 solution. But hold on, we need to convert that concentration from Molarity to moles. So, 71.4 mL * 0.150 mol/L = 10.71 moles of HNO3. Finally, we can convert that to volume by dividing by the concentration: 10.71 moles / 0.150 mol/L = 71.4 mL. Ta-da! So, the volume of HNO3 solution required is 71.4 mL.
B. Now let's calculate the mass of carbon dioxide produced. Given that the balanced equation shows a 1:1 ratio between CO2 and HNO3, and we just calculated that we have 10.71 moles of HNO3, it means we also have 10.71 moles of CO2. Now we just need to convert that to grams, using the molar mass of CO2 (44.01 g/mol): 10.71 moles * 44.01 g/mol = 470.9 grams. Voilà!
To determine the volume of the HNO3 solution needed to completely react with the Na2CO3 solution, we can use the stoichiometry of the balanced chemical equation.
Step 1: Find the number of moles of Na2CO3.
Using the given concentration of the Na2CO3 solution and the volume, we can calculate the moles of Na2CO3.
moles of Na2CO3 = concentration x volume
moles of Na2CO3 = 0.108 M x 35.7 mL = 3.8696 mmol
Step 2: Determine the number of moles of HNO3 required for the reaction.
From the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and HNO3 is 1:2.
So, moles of HNO3 = 2 x moles of Na2CO3
moles of HNO3 = 2 x 3.8696 mmol = 7.7392 mmol
Step 3: Calculate the volume of the HNO3 solution.
To convert the moles of HNO3 to volume, we need to use its concentration.
moles of HNO3 = concentration x volume
7.7392 mmol = 0.150 M x volume
volume = 7.7392 mmol / 0.150 M
volume = 51.6 mL
Rounding to three significant figures, the volume of the HNO3 solution required is 51.4 mL (Answer A).
Now, let's move on to part B to determine the mass of carbon dioxide (CO2) formed in the reaction.
Step 1: Find the number of moles of CO2.
From the balanced equation, we can see that the stoichiometric ratio between Na2CO3 and CO2 is 1:1.
So, the moles of CO2 formed will be the same as the moles of Na2CO3 used.
moles of CO2 = 3.8696 mmol
Step 2: Convert the moles of CO2 to grams.
To convert moles to grams, we need to use the molar mass of CO2, which is 44.01 g/mol.
mass of CO2 = moles of CO2 x molar mass of CO2
mass of CO2 = 3.8696 mmol x 44.01 g/mol
mass of CO2 = 170.22 mg
Converting mg to grams:
mass of CO2 = 170.22 mg / 1000 = 0.17022 g
Therefore, the mass of carbon dioxide formed in the reaction is 0.17022 grams (Answer B).
To find the volume of HNO3 solution required for the reaction, we will use the concept of stoichiometry and the given balanced chemical equation.
First, let's calculate the number of moles of Na2CO3 using its molarity and volume.
moles of Na2CO3 = molarity × volume
moles of Na2CO3 = 0.108 M × 35.7 mL
moles of Na2CO3 = 0.108 mol/L × 0.0357 L
moles of Na2CO3 = 0.0038616 mol (rounding to 4 significant figures)
According to the balanced chemical equation, the stoichiometric ratio between Na2CO3 and HNO3 is 1:2. Therefore, the number of moles of HNO3 required will be twice that of Na2CO3.
moles of HNO3 = 2 × moles of Na2CO3
moles of HNO3 = 2 × 0.0038616 mol
moles of HNO3 = 0.0077232 mol (rounding to 4 significant figures)
Next, let's calculate the volume of HNO3 solution using its molarity and the calculated number of moles of HNO3.
volume of HNO3 solution = moles of HNO3/molarity
volume of HNO3 solution = 0.0077232 mol ÷ 0.150 mol/L
volume of HNO3 solution = 0.0514827 L
volume of HNO3 solution = 51.5 mL (rounding to three significant figures)
So, the volume of the 0.150 M HNO3 solution required for the reaction is 51.5 mL.
Now, let's move on to part B to find the mass of carbon dioxide (CO2) formed in the reaction.
According to the balanced chemical equation, we can see that for every 2 moles of HNO3, we produce 1 mole of CO2. Therefore, the number of moles of CO2 formed will be half that of the moles of HNO3 used.
moles of CO2 = 0.5 × moles of HNO3
moles of CO2 = 0.5 × 0.0077232 mol
moles of CO2 = 0.0038616 mol (rounding to 4 significant figures)
To determine the mass of CO2 formed, we need to use its molar mass, which is approximately 44.01 g/mol.
mass of CO2 = moles of CO2 × molar mass of CO2
mass of CO2 = 0.0038616 mol × 44.01 g/mol
mass of CO2 = 0.1702 g (rounding to 4 significant figures)
Therefore, the mass of carbon dioxide formed in the reaction is approximately 0.1702 g.