Respuesta :
Answer:
The correct answer is 4.46 L.
Explanation:
Based on the given information, the mass of NiO given is 23.87 grams. The no. of moles can be determined by using the formula, n = weight/molecular mass. The molecular mass of NiO is 74.69 gram per mole. Now the moles of NiO will be,
Moles of NiO = 23.87 g/74.69 g/mol = 0.3196 moles
Now the moles of O2 will be = 1/2 * moles of NiO
= 1/2 * 0.3196 = 0.160 moles
The total pressure given in the question is 755 mmHg, and the vapor pressure mentioned is 55.4 mmHg. So, the pressure of O2 will be,
= Total pressure - Vapor pressure
= 755 mmHg - 55.4 mmHg
= 699.6 mmHg
= 699.6 * (1 atm / 760 mmHg)
= 0.9205 atm
Now the volume of the oxygen gas collected can be calculated by using the ideal gas law equation, that is, PV = nRT
0.9205 atm . V = 0.160 moles * 0.08206 L-atm/mol-K * 313 K (273 + 40 K)
V = 0.160 moles * 0.08206 L-atm/mol-K * 313 K / 0.9205 atm
V = 4.46 L
The volume of the gas collected after the reaction have been 4.46L.
Moles can be defined as the mass per unit molecular mass.
Moles = [tex]\rm \dfrac{weight}{molecular\;weight}[/tex]
Moles of NiO = [tex]\rm \dfrac{23.87\;g}{74.69\;g/mol}[/tex]
Moles of NiO = 0.3196 mol.
The moles of oxygen produced have been half the moles of NiO.
Moles of oxygen = 0.160 moles.
Total pressure = Vapor pressure of NiO + Vapor pressure of Oxygen
755 mm Hg = Vapor pressure of NiO + 55.4 mm Hg
Vapor pressure of NiO = 699.6 mmHg
699.6 mmHg = 0.9205 atm.
From the ideal equation:
PV = nRT
0.9205 atm × Volume = 0.160 × 0.08206 × 313 K
Volume = 4.46L.
The volume of the gas collected after the reaction have been 4.46L.
For more information about vapor pressure, refer to the link:
https://brainly.com/question/17757365
