if 50cm3 of Co2 were put in 40cm3 of O2. Determine the gas that would be in excess and calculate the volume of that would be unused
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In order to determine the gas that would be in excess and calculate the volume that would be unused, we need to compare the stoichiometric ratio of CO2 to O2.
The balanced chemical equation for the reaction between CO2 and O2 is:
2CO2 + O2 → 2CO + O2
From the balanced equation, we can see that for every 2 molecules of CO2, 1 molecule of O2 is required to completely react.
Given that we have 50 cm3 of CO2 and 40 cm3 of O2, we can calculate the number of moles for each gas using the ideal gas law, assuming the temperature and pressure are constant.
To calculate the number of moles, we need to use the molar volume of gases at standard temperature and pressure (STP), which is 22.4 L/mol.
Number of moles of CO2 = Volume (in liters) / Molar volume
= 50 cm3 / 1000 cm3/L
= 0.05 L / 22.4 L/mol
= 0.002232 moles
Number of moles of O2 = Volume (in liters) / Molar volume
= 40 cm3 / 1000 cm3/L
= 0.04 L / 22.4 L/mol
= 0.001786 moles
Now, let's compare the stoichiometric ratio to determine the gas in excess.
Since the ratio of CO2 to O2 is 2:1 in the balanced equation, we need to divide the moles of each gas by their respective coefficients in the balanced equation.
Moles of CO2 needed = 0.001786 moles / 1
= 0.001786 moles
Moles of O2 needed = 0.001786 moles / 2
= 0.000893 moles
Comparing the moles of CO2 available (0.002232 moles) to the moles needed (0.001786 moles), we can see that the amount of CO2 available is greater than the amount needed. Therefore, CO2 would be in excess.
To calculate the volume of gas that would be unused, we need to convert the number of moles of the gas in excess back to volume.
Volume of CO2 unused = Moles of CO2 in excess x Molar volume
= (0.002232 - 0.001786) moles x 22.4 L/mol
= 0.000446 moles x 22.4 L/mol
= 0.009984 L
= 9.984 cm3
Therefore, the volume of CO2 that would be unused is approximately 9.984 cm3.
To determine the gas that would be in excess and calculate the volume of that gas that would be unused, we first need to compare the stoichiometry (molar ratio) of the balanced chemical equation between CO2 and O2. However, since you only provided the volumes of gases, we need to convert those volumes into moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (assumed constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (assumed constant)
Since the pressure, gas constant, and temperature are assumed constant and are not provided in the question, we can ignore them for now.
To convert volumes into moles, we need to know the molar volume of each gas at the given conditions. The molar volume is defined as the volume occupied by one mole of a gas at a specific temperature and pressure. For simplicity, we'll assume standard temperature and pressure (STP), which is 273.15 K and 1 atm.
At STP, the molar volume is approximately 22.4 L/mol for any gas.
Now let's calculate the number of moles for each gas:
CO2:
50 cm^3 = 50/1000 L (converting cm^3 to L)
= 0.05 L
n(CO2) = 0.05 L / 22.4 L/mol
≈ 0.00223 mol
O2:
40 cm^3 = 40/1000 L (converting cm^3 to L)
= 0.04 L
n(O2) = 0.04 L / 22.4 L/mol
≈ 0.00179 mol
According to the balanced chemical equation, the stoichiometry between CO2 and O2 is 1:2. This means that for every 1 mole of CO2, we need 2 moles of O2.
Since we have 0.00223 moles of CO2 and 0.00179 moles of O2, we can multiply the number of moles of O2 by the stoichiometric coefficient ratio to determine the required moles of O2:
Required moles of O2 = 2 × (number of moles of CO2)
= 2 × 0.00223 mol
≈ 0.00446 mol
Comparing this with the actual moles of O2 (0.00179 mol), we see that we have less O2 than required. Therefore, O2 is the limiting reactant, and CO2 will be in excess.
To calculate the volume of unused CO2, we can convert the moles of CO2 to volume using the molar volume at STP:
Unused volume of CO2 = moles of CO2 × molar volume at STP
= 0.00223 mol × 22.4 L/mol
≈ 0.05 L
≈ 50 cm^3
Therefore, the volume of CO2 that would be unused is approximately 50 cm^3.