The question:
Formic acid when warmed with sulfuric acid, decomposes to water and carbon monoxide.
HCHO2(l)-->H2O(l)+CO(g)
If 3.85L of carbon monoxide was collected over water at 25C and 589mmHg, how many grams of formic acid were consumed?
(This problem requires that you recognize the partial pressure of water is included in the 689mmHg and must be accounted for). (answer: 6.34g)
Can someone please help me with this, my teacher does not explain this well and I have a bunch of these kinds of questions, hopefully with steps on how to solve one I can figure the rest out?
DISREGARD
To solve this problem, we need to use the ideal gas law and consider the partial pressure of water vapor. Here's a step-by-step guide on how to find the number of grams of formic acid consumed:
Step 1: Convert the given conditions to the proper units
- Temperature: 25°C = 298 K.
- Pressure: 689 mmHg (total pressure) - The partial pressure of water vapor needs to be subtracted.
- Volume: 3.85 L.
Step 2: Calculate the partial pressure of water vapor using Dalton's law of partial pressures.
The partial pressure of water vapor at 25°C is 23.8 mmHg (you can find this value in a reference table). Subtracting this from the total pressure gives us the partial pressure of carbon monoxide: 689 mmHg - 23.8 mmHg = 665.2 mmHg.
Step 3: Convert the partial pressure of carbon monoxide to atmospheres.
Since the ideal gas law requires pressure in atm, we need to convert the partial pressure of CO to atm by dividing by 760 (1 atm = 760 mmHg): 665.2 mmHg / 760 = 0.8737 atm.
Step 4: Calculate the number of moles of CO produced.
Use the ideal gas law equation: PV = nRT.
- P = pressure in atm (0.8737 atm).
- V = volume in liters (3.85 L).
- R = ideal gas constant (0.0821 L·atm/(mol·K)).
- T = temperature in Kelvin (298 K).
Rearranging the equation to solve for moles (n), we get:
n = (PV) / RT = (0.8737 atm) * (3.85 L) / (0.0821 L·atm/(mol·K) * 298 K).
Solving this equation, we find that the number of moles of CO produced is 0.1396 mol.
Step 5: Determine the stoichiometry of the reaction.
The balanced equation is: HCHO2(l) --> H2O(l) + CO(g).
From the equation, we can see that 1 mol of HCHO2 reacts to produce 1 mol of CO.
Step 6: Calculate the number of moles of HCHO2 consumed.
Since the stoichiometry is 1:1, the number of moles of HCHO2 consumed is also 0.1396 mol.
Step 7: Convert moles of HCHO2 to grams of HCHO2.
To calculate the mass, we need to use the molar mass of HCHO2. The molar mass of HCHO2 is approximately 46 g/mol.
Mass = moles * molar mass = 0.1396 mol * 46 g/mol = 6.4096 g = 6.41 g (rounded to two decimal places).
Therefore, approximately 6.34 grams of formic acid were consumed in the reaction.
Remember to double-check your calculations and significant figures along the way.