calculate the volume of Nitrogen gas generated at 21 degree celcius and 823 mmHg by the decomposition of 60.0 grams of sodium nitrate?

Heating NaNO3 produces O2 and not N2. Did you mean N2? If so, then

2NaNO3 ==> 2NaNO2 + O2
mols NaNO3 = grams NaNO3/molar mass = 60/approx 85 = about 0.7
Since 2 mol NaNO3 will produce 1 mol O2, then 0.7 mol NaNO3 will produce 0.7/2 = approx 0.35 mol O2.
Then use PV = nRT so solve for the volume.
NOTE: I have estimated all of the numbers. They are close. You need to do them more accurately. Post your work if you have trouble.

That was my comment also: what do you mean by decomposition? real world, or some fake chemical decomposition given as a teacher thinking problem. Sodium nitrite is very stable, and it (to my knowledge) does not decompose.

To calculate the volume of nitrogen gas generated, you would need to use the ideal gas law equation: 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.

First, you need to determine the number of moles of nitrogen gas generated using the given mass of sodium nitrate. The molar mass of sodium nitrate (NaNO3) is 85 grams/mol.

Number of moles = Mass / Molar mass
Number of moles = 60.0 g / 85 g/mol
Number of moles = 0.706 moles

Next, you need to convert the pressure to atmospheres. The conversion factor is 1 atm = 760 mmHg.

Pressure = 823 mmHg / 760 mmHg/atm
Pressure = 1.083 atm

Now, we need to convert the temperature from Celsius to Kelvin. The conversion formula is Kelvin = Celsius + 273.15.

Temperature in Kelvin = 21 °C + 273.15
Temperature in Kelvin = 294.15 K

Now we can substitute the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P
V = (0.706 moles * 0.0821 L•atm/mol•K * 294.15 K) / 1.083 atm

V ≈ 16.57 liters

Therefore, the volume of nitrogen gas generated by the decomposition of 60.0 grams of sodium nitrate at 21°C and 823 mmHg is approximately 16.57 liters.

To calculate the volume of nitrogen gas generated, we need to use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles (in mol)
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

First, let's convert the given temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15 = 21 + 273.15 = 294.15 K

Next, we need to calculate the number of moles of nitrogen gas generated. From the balanced equation for the decomposition of sodium nitrate:

2NaNO3 → 2NaNO2 + O2

we can see that for every 2 moles of sodium nitrate, we get 1 mole of nitrogen gas. So, we first need to find the number of moles of sodium nitrate.

Molar mass of NaNO3:
Na = 22.99 g/mol
N = 14.01 g/mol
O = 16.00 g/mol

Total molar mass = (22.99 g/mol + 14.01 g/mol + 3 * 16.00 g/mol) = 85.00 g/mol

Now, we can calculate the number of moles of sodium nitrate:
n = mass / molar mass = 60.0 g / 85.00 g/mol = 0.706 mol

Since 2 moles of sodium nitrate yield 1 mole of nitrogen gas, the number of moles of nitrogen gas is also 0.706 mol.

Finally, we can now use the ideal gas law to calculate the volume of nitrogen gas generated:

PV = nRT

V = nRT / P
= (0.706 mol)(0.0821 L.atm/mol.K)(294.15 K) / (823 mmHg / 760 mmHg/atm)
≈ 0.160 L (rounded to three decimal places)

Therefore, the volume of nitrogen gas generated is approximately 0.160 liters.