What will be the volume of Co2 at NTP obtain by heating 16.8gm NaHCo3
how many moles NaHCO3 in 16.8 g?
each mole occupies 22.4 L, right?
To find the volume of CO2 at NTP (Normal Temperature and Pressure) generated by heating 16.8 grams of NaHCO3 (sodium bicarbonate), we need to use the ideal gas law equation.
The ideal gas law equation is as follows:
PV = nRT
Where:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
First, we need to determine the number of moles (n) of NaHCO3. We can do this by using the molar mass of NaHCO3.
The molar mass of NaHCO3:
Na = 22.99 g/mol
H = 1.01 g/mol (in H2CO3)
C = 12.01 g/mol
O = 16.00 g/mol (in H2CO3)
NaHCO3 molar mass = 22.99 + 1.01 + 12.01 + (3 × 16.00) = 84.01 g/mol
Next, we can calculate the number of moles (n):
n = mass / molar mass
n = 16.8 g / 84.01 g/mol
Now, we have the number of moles (n) of NaHCO3. Since one mole of NaHCO3 produces one mole of CO2, the number of moles of CO2 produced will be the same.
To convert moles to volume, we will use the ideal gas law equation. At NTP, the temperature (T) will be 273.15 K, and the pressure (P) will be 1 atm.
Solving the ideal gas law equation for volume (V), we get:
V = (nRT) / P
Substituting the known values, we can calculate the volume of CO2:
V = (n × R × T) / P
V = (n × 0.0821 L·atm/(mol·K) × 273.15 K) / 1 atm
By substituting the calculated value of n, you can find the volume of CO2 generated by heating 16.8 grams of NaHCO3 at NTP.
No. 1 mol of gas occupies 22.4 L @ STP but 1 mol of gas occupies 24.0 L @ NTP (STP uses 273 for T and NTP uses 293 for T).
Also note that 1 mol NaHCO3 produces either 1 mol CO2 or1/2 mol CO2 depending upon the degradation of the NaHCO3.
2NaHCO3 ==> Na2CO3 + CO2 + H2O or
2NaHCO3 ==> Na2O + 2CO2 + H2O