given 2CH3OH+3O2->2CO2+4H2O
1. Calculate number of grams of CH3OH required to react with 34.0L of O2 at STP.
2. How many liters of CO2 at 18 degrees Celcius and 6.94 atm are produced when 3.40 mol of CH3Oh is burned completely.
3. What volume of CO2 is produced if 7.11L of O2 at STP reacts with an excess of liquid CH3OH and STP conditions are restored.
Show all work involved.
2CH3OH + 3O2 ==> 2CO2 + 4H2O
1. You know 1 mol O2 at STP will occupy 22.4L. Convert 34.0L O2 to mols. mols = 34.0/22.4 = 1.517 which I would round to 1.52 mols.
Using the coefficients in the balanced equation, convert mols O2 to mols CH3OH. That's 1.52 mols O2 x (2 mols CH3OH/3 mols O2) = 1.01 mols CH3OH.
Now convert mols CH3OH to grams. g = 1.01 x 32 = 32.4 g CH3OH.
The other parts are done the same way. Post your work if you get stuck.
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1. To calculate the number of grams of CH3OH required to react with 34.0L of O2 at STP (Standard Temperature and Pressure), you need to use the ideal gas law equation. The ideal gas law equation is given by PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
At STP, the pressure is 1 atm and the temperature is 273.15 K.
The equation for the reaction of CH3OH with O2 is 2CH3OH + 3O2 → 2CO2 + 4H2O.
First, we need to find the number of moles of O2 using the ideal gas law. We can rearrange the ideal gas law equation as follows: n = PV / RT.
Given:
Pressure (P) = 1 atm
Volume (V) = 34.0 L
Temperature (T) = 273.15 K (STP)
Ideal Gas Constant (R) = 0.0821 L·atm/(mol·K)
n = (1 atm * 34.0 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = 1.317 mol of O2
Using the stoichiometry of the balanced chemical equation, we can determine the moles of CH3OH required. From the balanced equation, we know that 2 moles of CH3OH react with 3 moles of O2.
Moles of CH3OH = (2/3) * n
Moles of CH3OH = (2/3) * 1.317 mol
Moles of CH3OH = 0.878 mol
To convert moles to grams, we need to use the molar mass of CH3OH (methanol). The molar mass of CH3OH is calculated by adding up the atomic masses of each element in the compound: (12.01 g/mol * 1) + (1.01 g/mol * 4) + (16.00 g/mol) = 32.04 g/mol.
Grams of CH3OH = Moles of CH3OH * Molar mass of CH3OH
Grams of CH3OH = 0.878 mol * 32.04 g/mol = 28.10 g
Therefore, approximately 28.10 grams of CH3OH are required to react with 34.0 liters of O2 at STP.
2. To calculate the number of liters of CO2 produced when 3.40 mol of CH3OH is burned completely, we need to use the balanced chemical equation and stoichiometry.
From the balanced equation, we know that 2 moles of CH3OH produce 2 moles of CO2.
Moles of CO2 = (2/2) * 3.40 mol
Moles of CO2 = 3.40 mol
To calculate the volume of CO2 at 18°C and 6.94 atm, we can use the ideal gas law equation: PV = nRT. Rearranging the equation, we have V = (nRT) / P, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature.
Given:
Pressure (P) = 6.94 atm
Volume (V) = ?
Temperature (T) = 18°C = 18 + 273.15 = 291.15 K
Ideal Gas Constant (R) = 0.0821 L·atm/(mol·K)
V = (3.40 mol * 0.0821 L·atm/(mol·K) * 291.15 K) / 6.94 atm
V = 0.340 L
Therefore, approximately 0.340 liters of CO2 are produced when 3.40 mol of CH3OH is burned completely at 18 degrees Celsius and 6.94 atm.
3. To calculate the volume of CO2 produced when 7.11 L of O2 at STP reacts with an excess of liquid CH3OH and under STP conditions, we first need to find the number of moles of O2 using the ideal gas law.
Using the ideal gas law equation, n = PV / RT, plugging in the values:
Pressure (P) = 1 atm
Volume (V) = 7.11 L
Temperature (T) = 273.15 K (STP)
Ideal Gas Constant (R) = 0.0821 L·atm/(mol·K)
n = (1 atm * 7.11 L) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = 0.307 mol of O2
From the balanced equation, we know that 3 moles of O2 produce 2 moles of CO2.
Moles of CO2 = (2/3) * n
Moles of CO2 = (2/3) * 0.307 mol
Moles of CO2 = 0.205 mol
Since the reaction is at STP, the volume of CO2 produced can be calculated using the ideal gas law equation: V = nRT / P.
Pressure (P) = 1 atm
Volume (V) = ?
Temperature (T) = 273.15 K (STP)
Ideal Gas Constant (R) = 0.0821 L·atm/(mol·K)
Moles of CO2 = 0.205 mol
V = (0.205 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm
V = 4.60 L
Therefore, approximately 4.60 liters of CO2 are produced when 7.11L of O2 at STP reacts with an excess of liquid CH3OH, and STP conditions are restored.
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