Carbon monoxide burns in oxygen to form carbon dioxide. what volume of oxygen at 25 degrees Celsius and 760. mmHg is required to react with 15.0 L of carbon monoxide which id at STP?
2CO + O2 ==> 2CO2
15.0 L CO at STP = 15.0/22.4 = ?mols
Convert mols CO to mol O2, then use PV = nRT at the conditions listed to solve for volume (in liters) of O2.
To find the volume of oxygen required to react with 15.0 L of carbon monoxide at STP, we can use the stoichiometry of the balanced chemical equation for the reaction between carbon monoxide (CO) and oxygen (O2). The balanced equation is:
2CO + O2 -> 2CO2
From the equation, we can see that the ratio of CO to O2 is 2:1. This means that for every 2 moles of carbon monoxide, we need 1 mole of oxygen.
Step 1: Convert the given volume of carbon monoxide at STP to moles.
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Therefore, 15.0 L of carbon monoxide is equal to 15.0/22.4 = 0.6696 moles of CO.
Step 2: Determine the moles of oxygen required.
Since the ratio of CO to O2 is 2:1, for every 2 moles of CO, we need 1 mole of O2. Therefore, we need half the moles of CO, which is 0.6696/2 = 0.3348 moles of O2.
Step 3: Convert moles of oxygen to volume at 25 degrees Celsius and 760 mmHg.
To convert moles of O2 to volume, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (760 mmHg)
V = volume (we need to solve for this)
n = number of moles (0.3348 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25°C + 273 = 298 K)
We rearrange the equation to solve for V:
V = nRT/P
V = (0.3348 mol)(0.0821 L·atm/(mol·K))(298 K)/(760 mmHg)
Step 4: Convert mmHg to atm and Celsius to Kelvin.
1 atm = 760 mmHg
To convert mmHg to atm, divide by 760:
V = (0.3348 mol)(0.0821 L·atm/(mol·K))(298 K)/(760/760)
Final Answer:
V ≈ 0.0418 L (rounded to four decimal places)
Therefore, at 25 degrees Celsius and 760 mmHg, approximately 0.0418 L of oxygen is required to react with 15.0 L of carbon monoxide at STP.
To solve this problem, we need to use the ideal gas law equation, which states:
PV = nRT
Where:
P = Pressure (in this case, given as 760. mmHg)
V = Volume (in this case, unknown)
n = Number of moles of gas (unknown)
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature (in this case, given as 25 degrees Celsius or 298 K)
First, we need to find the number of moles of carbon monoxide using the volume provided at STP (Standard Temperature and Pressure). STP conditions refer to a temperature of 273 K (0 degrees Celsius) and a pressure of 1 atmosphere (atm).
We can use the ideal gas law equation at STP to find the number of moles:
PV = nRT
(1 atm) * (15.0 L) = n * (0.0821 L·atm/mol·K) * (273 K)
Now, we can solve for n:
n = (1 atm * 15.0 L) / (0.0821 L·atm/mol·K * 273 K)
n ≈ 0.607 mol
Now, we need to find the number of moles of oxygen required to react with 0.607 mol of carbon monoxide. The balanced equation shows that the ratio of carbon monoxide to oxygen is 1:1. This means that we need an equal number of moles of oxygen.
Therefore, we need 0.607 mol of oxygen.
Now, we can find the volume of oxygen needed at 25 degrees Celsius and 760. mmHg using the ideal gas law equation:
PV = nRT
(760. mmHg) * (V) = (0.607 mol) * (0.0821 L·atm/mol·K) * (298 K)
Now, we can solve for V:
V = (0.607 mol * 0.0821 L·atm/mol·K * 298 K) / (760. mmHg)
V ≈ 0.177 L
Therefore, approximately 0.177 L of oxygen at 25 degrees Celsius and 760. mmHg is required to react with 15.0 L of carbon monoxide at STP.