Answer :
The number of moles of NH4NO2 that reacted is:
10.74 moles / 2 = 5.37 moles
Approximately 343.29 grams of ammonium nitrite must have reacted.
The number of moles of Fe2(CO3)3 available is the limiting reactant since it is less than half the moles of H3PO4. Thus, Fe2(CO3)3 is the limiting reactant.
To determine the number of grams of ammonium nitrite that reacted, we first need to calculate the number of moles of nitrogen gas using the ideal gas law equation:
PV = nRT
Given:
Volume (V) = 2.58 L
Temperature (T) = 21.0°C + 273.15 = 294.15 K
Pressure (P) = 97.8 kPa = 97.8 * 101.325 Pa (since 1 kPa = 101.325 Pa)
Now, rearranging the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
n = (97.8 * 101.325 Pa) * (2.58 L) / (8.314 J/(mol·K) * 294.15 K)
n ≈ 10.74 moles of nitrogen gas
From the balanced equation:
NH4NO2 -> N2 + 2H2O
It can be observed that 2 moles of NH4NO2 produce 1 mole of N2.
Therefore, the number of moles of NH4NO2 that reacted is:
10.74 moles / 2 = 5.37 moles
To determine the mass of NH4NO2, we need to use its molar mass, which is:
(14.01 g/mol + 1.01 g/mol) + (14.01 g/mol + 16.00 g/mol + 16.00 g/mol) = 64.05 g/mol
Mass of NH4NO2 = 5.37 moles * 64.05 g/mol ≈ 343.29 grams
Therefore, approximately 343.29 grams of ammonium nitrite must have reacted.
Now, moving on to the remaining questions:
If the experiment is done at a significantly higher temperature, the volume of nitrogen gas will increase. According to Charles's Law, the volume of a gas is directly proportional to its temperature, assuming constant pressure and amount of gas. As the temperature increases, the gas molecules gain more kinetic energy and move with greater speed, leading to increased collisions and expansion of the gas volume.
B. If the amount of ammonium nitrite is increased, the volume of nitrogen gas produced will remain the same. The balanced equation shows that the stoichiometry of the reaction is 1:1 between NH4NO2 and N2. Therefore, increasing the amount of ammonium nitrite will only result in a proportional increase in the amount of nitrogen gas produced, maintaining the same volume.
C. If the experiment is not collected over water, the volume of nitrogen gas will remain the same. Collecting the gas over water in a gas collecting tube accounts for the water vapor pressure and helps to measure the pure volume of nitrogen gas. If the gas is collected without considering the water vapor pressure, the total volume of the collected gas would include the volume occupied by water vapor, but the volume of nitrogen gas itself would remain the same.
To determine the limiting reactant, we need to compare the number of moles of each reactant and the stoichiometry of the balanced equation.
Given:
Volume of phosphoric acid (H3PO4) = 900.0 mL = 0.9000 L
Molarity of H3PO4 = 3.00 M
Number of moles of H3PO4 = Molarity * Volume = 3.00 mol/L * 0.9000 L = 2.70 moles
Molar mass of Fe2(CO3)3 = 2 * (55.85 g/mol) + 3 * (12.01 g/mol + 16.00 g/mol + 16.00 g/mol) = 291.72 g/mol
Number of moles of Fe2(CO3)3 = Mass / Molar mass = 235 g / 291.72 g/mol ≈ 0.805 moles
According to the balanced equation:
Fe2(CO3)3 + 2H3PO4 -> 2FePO4 + 3H2O + 3CO2
From the stoichiometry, we can see that 1 mole of Fe2(CO3)3 reacts with 2 moles of H3PO4. Therefore, the number of moles of Fe2(CO3)3 available is the limiting reactant since it is less than half the moles of H3PO4. Thus, Fe2(CO3)3 is the limiting reactant.
For more question on moles
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