How do you do this? please help. Thank you.
A.) Suppose 2.5 grams of sodium hydrogen carbonate reacts with acetic acid, calculate how many liters of carbon dioxide gas will be produced at 22 degrees celcius and 1.10 atm.
B.) Suppose 0.990 liters of carbon dioxide at 25 degrees celcius and 755 mmHg is produced when sodium hydrogen carbonate reacts with acetic acid. Calculate how many grams of sodium hydrogen carbonate were used to start this reaction.
Also, would you have to use PV=nRT.
for A.) n and v was missing so how do I manipulate it to find both. Is there another equation perhaps? for part B too.
First write the equation of the reaction:
reactants: NaHCO3+CH3COOH
products: CH3COONa+CO2+H2O
From the molar masses of
NaHCO3 = 23+1+12+3*16 = 84
CO2 = 12+2*16 = 44
We find the stoichiometric ratio of the two compounds to be
84:44 (approximately)
Hence by proportions/ratios, we find the number of grams, and moles of CO2 produced.
Yes, you need to convert the volume to 22° C and 1.10 atm using PV=nRT, with temperatures being in °K.
Part B is essentially the same as part A, but backwards.
To solve problem A, we can use the ideal gas law equation, which is 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, we need to convert the given temperature from Celsius to Kelvin by adding 273.15. So, T = 22 + 273.15 = 295.15 K.
Next, we can rearrange the ideal gas law equation to solve for the number of moles. The equation becomes n = PV / RT.
We are given the pressure, P = 1.10 atm, and we need to find the volume, V. However, the volume is not provided directly. We only have the mass of sodium hydrogen carbonate, which is not enough to find the volume.
To calculate the volume, we need to use the molar mass of sodium hydrogen carbonate (NaHCO3) and the mass given.
The molar mass of NaHCO3 is:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 atoms)
Adding these values, we find the molar mass of NaHCO3 is 84.01 g/mol.
Given that you have 2.5 grams of NaHCO3, you can convert this to moles using the equation:
moles = mass / molar mass
moles = 2.5 g / 84.01 g/mol
Now that you have the moles of NaHCO3, you can use the ideal gas law equation to find the volume:
V = nRT / P
V = (moles of NaHCO3) * R * T / P
Plug in the values for moles (from the previous calculation), R (the ideal gas constant = 0.0821 L atm / (K mol)), T (295.15 K), and P (1.10 atm) into the equation to find the volume.
For problem B, you are given the volume, V = 0.990 L, and the pressure, P = 755 mmHg.
First, convert the pressure from mmHg to atm by dividing by 760 mmHg/atm:
P = 755 mmHg / 760 mmHg/atm = 0.9947 atm.
Next, convert the temperature from Celsius to Kelvin: T = 25 + 273.15 = 298.15 K.
Rearranging the ideal gas law equation, n = PV / RT, we can solve for the number of moles.
Now we can plug in the values for pressure, volume, R (0.0821 L atm / (K mol)), and T into the equation to find the number of moles.
Once we have the number of moles, we can use the molar mass of NaHCO3 (as calculated in problem A) to find the mass of sodium hydrogen carbonate using the equation: mass = moles * molar mass.
Finally, you will have the mass of NaHCO3 used to start the reaction.
Remember to check your units and ensure they are consistent throughout the calculations.