2NiO(s)→2Ni(s)+O2(g)
If O2 is collected over water at 40.0C and a total pressure of 739mmHg , what volume of gas will be collected for the complete reaction of 24.18g of NiO?
mols NiO = 24.18 g NiO/molar mass NiO.
Convert mols NiO to mols O2 using the coefficients in the balanced equation.
Using PV = nRT convert n = mols to volume at the stated conditions.
For pressure, use 739-vapor pressure H2O at 40 C.
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's calculate the number of moles of O2 gas produced from the reaction:
2NiO(s) → 2Ni(s) + O2(g)
From the chemical equation, we can see that 2 moles of NiO react to produce 1 mole of O2. So, we need to find the number of moles of NiO first.
The molar mass of NiO can be calculated using the atomic masses of nickel (Ni) and oxygen (O) from the periodic table. The atomic mass of Ni is 58.69 g/mol, and the atomic mass of O is 16.00 g/mol. Therefore, the molar mass of NiO is:
NiO: (1 * Ni) + (1 * O) = (1 * 58.69 g/mol) + (1 * 16.00 g/mol) = 74.69 g/mol
Now, we can calculate the number of moles of NiO:
Moles of NiO = Mass of NiO / Molar mass of NiO
= 24.18 g / 74.69 g/mol
≈ 0.3238 mol
Since the reaction produces 1 mole of O2 for every 2 moles of NiO, the number of moles of O2 will be half the number of moles of NiO:
Moles of O2 = (1/2) * Moles of NiO
= (1/2) * 0.3238 mol
≈ 0.1619 mol
Now, we can use the ideal gas law to find the volume of O2 gas collected.
R, the ideal gas constant, is 0.0821 L·atm/(mol·K). Since the pressure is given in mmHg, we need to convert it to atm:
Pressure (atm) = 739 mmHg / 760 mmHg/atm
≈ 0.972 atm
It is also essential to convert the temperature from Celsius to Kelvin:
Temperature (K) = 40.0°C + 273.15
= 313.15 K
Now we can substitute the known values into the ideal gas law equation to solve for the volume:
V = (n * R * T) / P
= (0.1619 mol * 0.0821 L·atm/(mol·K) * 313.15 K) / 0.972 atm
≈ 6.76 L
Therefore, the volume of O2 gas collected is approximately 6.76 liters.