3. The amount of ammonium nitrite that must have reacted is 8.164 grams.
- A. If the experiment is done at a significantly higher temperature, the volume of nitrogen gas will increase.
- B. If the amount of ammonium nitrite is increased, the volume of nitrogen gas will increase.
- C. If the experiment is not collected over water, the volume of nitrogen gas will remain the same.
4a. The limiting reactant is Fe₂(CO₃)₃.
b. The volume of carbon dioxide gas that can be produced is 619 milliliters.
What mass of ammonium nitrite reacted?
First, the number of moles of nitrogen gas using the ideal gas law:
PV = nRT
where:
P is pressure (in atm) = 97.8 kPa = 97.8/101.3 = 0.965 atm
V is volume (in liters) = 2.58 L
n is number of moles
R is gas constant = 0.0821 L·atm/(mol·K)
T is temperature (in Kelvin) = 21.0°C + 273.15 = 294.15 K
Solving for n:
n = (PV) / (RT)
n = (0.965 atm * 2.58 L) / (0.0821 L·atm/(mol·K) * 294.15 K)
n ≈ 0.102 moles of nitrogen gas
The moles of ammonium nitrite that reacted from the balanced chemical equation:
NHâ‚„NOâ‚‚ → Nâ‚‚ + 2Hâ‚‚O
From the mole ratio above, one mole of ammonium nitrite reacts to produce one mole of nitrogen gas
Therefore, the moles of ammonium nitrite reacted is 0.102 moles.
So, the mass of ammonium nitrite that reacted is:
Mass = moles * molar mass
Mass = 0.102 moles * 80.04 g/mol
Mass ≈ 8.164 g
4a. Moles of H₃PO₄ = volume (in liters) * molarity
= 0.900 L * 3.00 mol/L
Moles of H₃PO₄ = 2.70 mol
Moles of Fe₂(CO₃)₃ = mass / molar mass
= 235 g / (2 * 55.85 g/mol + 3 * 12.01 g/mol + 9 * 16.00 g/mol)
= 235 g / 291.88 g/mol
Moles of Fe₂(CO₃)₃ = 0.804 mol
Therefore, Fe₂(CO₃)₃ is the limiting reactant.
b. From the balanced equation: 1 mol Fe₂(CO₃)₃ produces 3 moles of CO₂
Moles of COâ‚‚ = 0.804 mol * 3
Moles of COâ‚‚ = 2.412 moles
the volume of COâ‚‚ will be:
PV = nRT
Where:
P is pressure = 45.5 psi or 3.099 atm
V is volume (in liters)
n is the number of moles = 2.412 mol
R is gas constant = 0.0821 L·atm/(mol·K)
T is temperature (in Kelvin) = 78°C + 273.15 = 351.15 K
Solving for V:
V = (nRT) / P
V = (2.412 mol * 0.0821 * 351.15/ 3.099
V ≈ 0.619 L = 619 mL
Learn more about limiting reactants at: https://brainly.com/question/30879855
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