14 grams of coal undergoes complete combustion in 58 grams of oxygen gas.
What volume of gas (measured at 25 degrees celsius and 100 kPa) is present when the reaction is complete?
This is a limiting reagent (LR) problem. I work these the long way but I think it's easier to explain this way.
C + O2 ==> CO2
mols C = grams/molar mass = 14/12 = approx 1.1 but you need to redo this for a more accurate number.
mols O2 = 58/32 = 1.8
Convert mols C to mols CO2. That's
approx 1.1 x (1 mol CO2/1 mol CO) = approx 1.1
Convert mols O2 to mols CO2. That's approx 1.8 x (1 mol CO2/1 mol O2) = 1.8
In LR problems the smaller amout is ALWAYS the correct answer and the reagent responsible for this is the LR. So C is the LR, O2 is the excess reagent. You want to know volume of gas. Some of that gas will be CO2 produced by the reaction BUT there will be some O2 not reacted and that will contribute to the total volume.
So how much O2 do we have that didn't react?
1.1 mols C x (1 mol O2/1 mol C) = 1.1 mols O2 used.
mols O2 not reacted is 1.8-1.1 = approx 0.7. So total mols gas is 1.1 from CO2 formed + 0.7 from the unreacted O2 = 1.8 total mols gas.
Now used PV = nRT and convert total mols to volume in liters. Post your work if you get stuck. REMEMBER these numbers I've used are estimates. You must go through and rework the problem with better numbers than I've used.
To answer this question, we need to use the concept of stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.
First, we need to determine the balanced chemical equation for the combustion of coal. Let's assume the coal is composed of carbon (C) only, and oxygen gas (O₂) is the reactant. The balanced equation for the combustion of carbon can be represented as:
C + O₂ -> CO₂
According to the equation, one mole of carbon reacts with one mole of oxygen gas to produce one mole of carbon dioxide (CO₂). To calculate the volume of gas produced, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (in kPa)
V = Volume (in liters)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/(mol·K))
T = Temperature (in Kelvin)
Now, let's find the number of moles of oxygen gas. The molar mass of O₂ is 32 g/mol, so:
Number of moles of O₂ = Mass of O₂ / Molar mass of O₂
= 58 g / 32 g/mol
= 1.8125 mol (rounded to 4 decimal places)
Since the balanced equation shows the stoichiometric ratio of 1 to 1 between oxygen gas and carbon, we can conclude that the number of moles of carbon is also 1.
Now, let's calculate the volume of gas at the given conditions (25 degrees Celsius and 100 kPa).
Note: We need to convert Celsius to Kelvin by adding 273.15.
T = 25 + 273.15
= 298.15 K
Using the ideal gas law equation, we can rearrange it to solve for volume (V):
V = (nRT) / P
V = (1.8125 mol * 0.0821 L·atm/(mol·K) * 298.15 K) / 100 kPa
V = 0.0469 L (rounded to 4 decimal places)
Therefore, the volume of gas produced when the combustion reaction is complete is approximately 0.0469 liters at 25 degrees Celsius and 100 kPa.