Barium carbonate and nitric acid reacts in a double displacement reaction. One of products also dissociates and evolves carbon dioxide. Write the net ionic reaction including the proper use of phase labels for this solution run in aqueous solution.
After this the following questions ask:
1. For the reaction in problem 2, if 25.0 mLs of 0.150M nitric acid was used, how many grams of barium nitrate would be required for complete reaction?
2. For the reaction from problem 2, if 0.5000 grams of barium carbonate was reacted with 25.0mLs of 0.15M nitric acid and the ags then collected in 100mL flask at 77 degrees celsius, what is the gas pressure in the flask?
**I am struggling big time on these questions. Please help! Thank you so muchhh**
BaCO3 + 2HNO3 ==> Ba(NO)3)2 + H2CO3 and
H2CO3 ==> H2O + CO2.
CO3^2-(aq) + 2H^+(aq) ==> H2O(l) + CO2(g) is the net ionic equation.
#1 makes no sense. Ba(NO3)2 is a PRODUCT, not a reactant.
#2.
BaCO3 + 2HNO3 ==> Ba(NO3)2 _+ H2O + CO2
mols BaCO3 = M x L = about 0.00375 but you should confirm that.
Convert mols BaCO3 to mols CO2
Then use PV = nRT. You know T, R, n, V, solve for P. Remember T must be in kelvin; P will be in atm if you use R = 0.08206.
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The balanced equation for the reaction between barium carbonate and nitric acid can be represented as follows:
BaCO3(s) + 2HNO3(aq) -> Ba(NO3)2(aq) + CO2(g) + H2O(l)
To write the net ionic equation, we need to separate the soluble ionic compounds into their respective ions. The molecular species that don't dissociate in solution are left in their molecular form. In this case, the net ionic equation is as follows:
Ba2+(aq) + CO3^2-(aq) + 2H+(aq) + 2NO3^-(aq) -> Ba2+(aq) + 2NO3^-(aq) + CO2(g) + H2O(l)
Now, let's proceed to answer the questions:
1. To find the number of grams of barium nitrate required for complete reaction, we need to determine the limiting reactant.
From the balanced equation, we can see that the molar ratio between barium carbonate and barium nitrate is 1:1. Thus, the moles of barium nitrate required will be equal to the moles of barium carbonate.
To calculate the moles of nitric acid used, we can use the formula:
moles = concentration × volume
moles of HNO3 = 0.150 M × 0.0250 L = 0.00375 moles
Since the stoichiometric ratio between nitric acid and barium carbonate is 2:1, the moles of barium carbonate required will be:
moles of BaCO3 = 0.00375 moles / 2 = 0.001875 moles
Now, to find the mass of barium nitrate, we can use its molar mass:
molar mass of Ba(NO3)2 = 137.33 g/mol
mass of Ba(NO3)2 = 0.001875 moles × 137.33 g/mol = 0.2578 grams
Therefore, approximately 0.2578 grams of barium nitrate would be required for complete reaction.
2. To calculate the gas pressure in the flask, we need to use the Ideal Gas Law. The formula is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
First, let's calculate the moles of carbon dioxide (CO2) produced from 0.5000 grams of barium carbonate:
moles of CO2 = mass / molar mass
moles of CO2 = 0.5000 g / 44.01 g/mol = 0.01136 moles
Next, we need to convert the volume of the flask from mL to liters:
Volume = 100 mL = 0.100 L
And convert the temperature from Celsius to Kelvin:
Temperature (K) = 77 °C + 273.15 = 350.15 K
Plugging in the values into the ideal gas law equation:
P × 0.100 L = 0.01136 moles × 0.0821 L·atm/mol·K × 350.15 K
Solving for P:
P = (0.01136 moles × 0.0821 L·atm/mol·K × 350.15 K) / 0.100 L
P ≈ 0.330 atm
Therefore, the gas pressure in the flask is approximately 0.330 atm.
To write the net ionic reaction for the reaction between barium carbonate (BaCO3) and nitric acid (HNO3), we first need to write the balanced chemical equation.
The balanced chemical equation for the reaction is as follows:
BaCO3 (s) + 2 HNO3 (aq) -> Ba(NO3)2 (aq) + H2O (l) + CO2 (g)
In this reaction, barium carbonate (BaCO3) reacts with nitric acid (HNO3) to form barium nitrate (Ba(NO3)2), water (H2O), and carbon dioxide (CO2).
To determine the amount of barium nitrate required for complete reaction, we can use the balanced chemical equation and stoichiometry.
1. We are given that 25.0 mL of 0.150 M nitric acid was used. First, convert the volume of nitric acid to moles:
Moles of HNO3 = Volume (L) x Concentration (M)
= 0.025 L x 0.150 M
= 0.00375 moles HNO3
2. According to the balanced chemical equation, 1 mole of BaCO3 reacts with 2 moles of HNO3. Therefore, the moles of BaCO3 required for complete reaction can be calculated:
Moles of BaCO3 = Moles of HNO3 / Stoichiometric ratio
= 0.00375 moles HNO3 / 2
= 0.001875 moles BaCO3
3. Now, we need to convert the moles of BaCO3 to grams. The molar mass of BaCO3 is 197.34 g/mol.
Mass of BaCO3 = Moles of BaCO3 x Molar mass of BaCO3
= 0.001875 moles BaCO3 x 197.34 g/mol
= 0.37 grams BaCO3
Therefore, 0.37 grams of barium carbonate would be required for complete reaction.
For the second question, we need to determine the gas pressure in the flask containing the collected gas at 77 degrees Celsius. To do this, we can use the ideal gas law:
PV = nRT
Where:
P = Pressure of the gas (in atm)
V = Volume of the flask (in L)
n = Number of moles of the gas
R = Ideal gas constant (0.0821 L atm/(mol K))
T = Temperature in Kelvin
1. We are given that the collected gas is carbon dioxide (CO2). First, we need to determine the number of moles of CO2 produced.
Moles of CO2 = Mass of CO2 / Molar mass of CO2
= 0.5000 g CO2 / 44.01 g/mol
= 0.01136 moles CO2
2. We are also given that the volume of the flask is 100 mL. Convert it to liters:
Volume of flask = 100 mL / 1000 mL/L
= 0.100 L
3. Convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 77 degrees Celsius + 273.15
= 350.15 K
Now, we can substitute the values into the ideal gas law equation:
P * 0.100 L = 0.01136 moles CO2 * 0.0821 L atm/(mol K) * 350.15 K
Solving for P:
P = (0.01136 moles CO2 * 0.0821 L atm/(mol K) * 350.15 K) / 0.100 L
= 0.0322 atm
Therefore, the gas pressure in the flask is 0.0322 atm.