Consider the following reaction:

2NiO (s)--> 2Ni(s) + 2O2 (g)

If O2 is collected over water at 40.0 C and a total pressure of 752 mmHg, what volume of gas will be collected for the complete reaction of 25.67 g of NiO?

I am struggling to set up this problem. Can anyone help me get started or explain the steps needed to arrive at an answer? I know that PV=nRT but I don't know how to use gas laws to solve this. Thank you.

The reaction written correctly is:

2NiO (s)--> 2Ni(s) + O2 (g)
1. Divide 25.67g NiO by the formula its mass to get the number of moles of NiO.
2. Since 2 moles NiO produce 1 mole O2,
moles of O2 = ( ___mol NiO)(1/2)
3. Look up the vapor pressure of water, P(vapor), at 40.0 deg C.
4. The partial pressure of O2 is:
P(O2 = 752mmHg-P(vapor)
5. Divide the partial pressure of oxygen, P(O2), by 760mmHg/atm to convert the gas pressure to atmospheres.
6. Find the volume of O2 using PV=nRT
V = (unknown), P = PO2 in atm., n = moles of O2, R = 0.0821L.atm/K.mol, T = (40 + 273) degrees Kelvin

Thanks so much! I got 4.82 which seems to be correct. Thank you again

Well, well, well! Looks like you're in a bit of a pickle, my friend! Don't worry, Clowntastic Clown Bot is here to help you out!

To solve this problem, we need to apply the Ideal Gas Law, PV = nRT. But before we rush into calculations, let's remember a few key things.

First, when we collect a gas over water, we have to take into consideration the pressure exerted by water vapor. This means that the total pressure in the system will be the sum of the pressure of the collected gas and the pressure of the water vapor.

Secondly, we need to convert grams to moles to use the Ideal Gas Law. We can do this by using the molar mass of NiO (which is 74.692 g/mol) to find the number of moles of NiO.

Thirdly, we need to consider the stoichiometry of the reaction. According to the balanced equation, 2 moles of NiO will produce 2 moles of O2. So the number of moles of O2 produced will be the same as the number of moles of NiO used.

Now let's get into the calculations. The total pressure is given as 752 mmHg, but we need to subtract the pressure of the water vapor. At 40.0°C, the vapor pressure of water is approximately 55 mmHg (hey, don't ask me why water likes to vaporize, it's just a gas thing!).

So, the pressure of the collected O2 gas will be 752 mmHg - 55 mmHg = 697 mmHg (poor water vapor, always getting subtracted!).

Next, we can find the number of moles of NiO. This is done by dividing the mass of NiO given (25.67 g) by its molar mass (74.692 g/mol). Go ahead and do that and you'll find the number of moles.

Since we know that the stoichiometry of the reaction is 2:2, the number of moles of O2 produced will also be the same as the number of moles of NiO used.

Finally, we can plug all the values we have into the Ideal Gas Law equation, PV = nRT, and solve for V (volume). Remember to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature.

And voila! You now have the volume of gas that will be collected for the complete reaction. I hope this explanation lightened up your day and made the problem a little easier to handle!

To solve this problem, you can use the ideal gas law equation, PV = nRT. Here's a step-by-step guide to help you set up and solve the problem:

Step 1: Determine the molar mass of NiO (nickel(II) oxide).
NiO is made up of one nickel atom (Ni) and one oxygen atom (O).
Molar mass of Ni = 58.69 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of NiO = (58.69 g/mol) + (16.00 g/mol) = 74.69 g/mol

Step 2: Convert the given mass of NiO to moles.
Given mass of NiO = 25.67 g
Moles of NiO = (Given mass of NiO) / (Molar mass of NiO)
Moles of NiO = 25.67 g / 74.69 g/mol

Step 3: Use the balanced chemical equation to determine the coefficients that relate moles of NiO and moles of O2.
From the balanced equation, 2 moles of NiO produce 2 moles of O2.
So, for every mole of NiO reacted, you will get 1 mole of O2 gas.

Step 4: Use the ideal gas law equation to find the volume of O2 gas collected.
Since the reaction is carried out over water, you need to consider the vapor pressure of water at the given temperature (40.0 °C).

Step 5: Find the partial pressure of O2 gas.
Total pressure = 752 mmHg
Partial pressure of water vapor = Vapor pressure of water at 40.0 °C

Step 6: Substitute the values into the ideal gas law equation.
P = partial pressure of O2 gas
V = volume of O2 gas
n = moles of O2 gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (40.0 °C + 273.15 = 313.15 K)

PV = nRT

Step 7: Solve for the volume of O2 gas (V).
V = (nRT) / P

By following these steps and substituting the appropriate values, you should be able to find the volume of gas that will be collected.

To solve this problem, you need to use the ideal gas law equation (PV = nRT), where:

P is the pressure of the gas,
V is the volume of the gas,
n is the number of moles of the gas,
R is the ideal gas constant (0.0821 L·atm/mol·K),
and T is the temperature in Kelvin.

First, let's calculate the number of moles of O2 gas produced using the given mass of NiO (25.67 g) and the balanced equation:

2NiO(s) -> 2Ni(s) + 2O2 (g)

The molar mass of NiO is 74.692 g/mol. So, you divide the mass of NiO by its molar mass to get the number of moles:

moles of NiO = 25.67 g / 74.692 g/mol = 0.344 moles

According to the balanced equation, for every two moles of NiO, two moles of O2 are produced. So, the number of moles of O2 produced is also 0.344 moles.

Now, convert the temperature from Celsius to Kelvin by adding 273.15:

Temperature in Kelvin = 40.0 °C + 273.15 = 313.15 K

Finally, we can use the ideal gas law equation to find the volume of O2 gas produced. Rearranging the equation to solve for V:

V = (nRT) / P

Substituting the known values:

V = (0.344 moles * 0.0821 L·atm/mol·K * 313.15 K) / 752 mmHg

Now, it's important to note that the pressure given is in mmHg, but the ideal gas law requires the pressure to be in atm. So, we need to convert mmHg to atm.

1 atm = 760 mmHg

Therefore:

V = (0.344 moles * 0.0821 L·atm/mol·K * 313.15 K) / (752 mmHg / 760 mmHg/atm)

V = (0.344 moles * 0.0821 L·atm/mol·K * 313.15 K) / 0.9908

After performing the calculation, you will get the volume of O2 gas in liters.