Volume-volume problems example what volume of O2 reacts CO to produce 0.5mole of Co2 at STP?
The balanced chemical equation for the reaction is:
2 CO + O2 -> 2 CO2
From the equation, we can see that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2. Therefore, to produce 0.5 mole of CO2, we need:
0.5 mol CO2 x (1 mol O2 / 2 mol CO) = 0.25 mol O2
Now we can use the ideal gas law to calculate the volume of O2 at STP (Standard Temperature and Pressure, 0°C and 1 atm) using the molar volume of a gas at STP, which is 22.4 L/mol:
PV = nRT
P = 1 atm
V = ?
n = 0.25 mol
R = 0.0821 L atm/mol K
T = 273 K
V = nRT/P = (0.25 mol)(0.0821 L atm/mol K)(273 K)/(1 atm) = 5.66 L
Therefore, the volume of O2 required to react with CO to produce 0.5 mole of CO2 at STP is 5.66 L.
what masses of CH4 formed when 8.3moles of CO react with H2
To solve this problem, we need to use the balanced chemical equation for the reaction between CO and H2 to form CH4:
CO + 3 H2 → CH4 + H2O
From the balanced equation, we see that 1 mole of CO reacts with 3 moles of H2 to produce 1 mole of CH4.
Therefore, if we have 8.3 moles of CO, we need 3 times that amount of H2 to completely react with the CO and produce CH4. So:
Number of moles of H2 = 8.3 mol CO x 3 mol H2 / 1 mol CO = 24.9 mol H2
Now we can calculate the limiting reactant. The limiting reactant is the reactant with the smallest number of moles compared to the stoichiometric ratio. The amount of CH4 formed will be determined by the limiting reactant.
To find the limiting reactant, we need to calculate the number of moles of CH4 that can be formed from each reactant, assuming that reactant is the limiting reactant.
From CO:
8.3 mol CO x 1 mol CH4 / 1 mol CO = 8.3 mol CH4
From H2:
24.9 mol H2 x 1 mol CH4 / 3 mol H2 = 8.3 mol CH4
Since both reactants produce an equal amount of CH4, neither reactant is limiting. Therefore, we can use either reactant to calculate the amount of CH4 formed.
From CO:
8.3 mol CO x 1 mol CH4 / 1 mol CO x 16.04 g CH4 / 1 mol CH4 = 133.9 g CH4
Therefore, 133.9 g of CH4 will be formed when 8.3 moles of CO react with excess H2.
To find the volume of O2 needed to react with CO, we can use the balanced chemical equation for the reaction between CO and O2:
2CO + O2 -> 2CO2
From the equation, we can see that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.
Since 0.5 moles of CO2 are produced, we can conclude that 0.25 moles of CO reacts with an equivalent amount of O2.
To calculate the volume of O2 needed, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm, which is 1 at STP)
V = volume (in liters, what we are trying to find)
n = number of moles
R = gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin, which is 273 K at STP)
Rearranging the equation to solve for V, we have:
V = (nRT) / P
Plugging in the values, we have:
V = (0.25 mol * 0.0821 L.atm/mol.K * 273 K) / 1 atm
V = 5.60 L
Therefore, the volume of O2 needed to produce 0.5 moles of CO2 at STP is 5.60 liters.