Toxic Carbon monoxide gas is produced when fossil fuels such as gasoline, burn wihtout sufficient oxygen gas. The CO ca eventually be converted to CO2 inthe atmsphere. Automobie cata;yctic converters are desgned to speed up this conversion:
How many moles of oxygen gas would be needed to convert 50.0 mol carbon monoxide to carbon dioxide?
What volume of oxygen gas would be needed to react with 986 L CO? (Assume both gases ar the same temp. and pressure)
2CO + O2 ==> 2CO2
50 moles CO gas will require 25 moles O2 gas.
986L CO x (1 mole O2/2 moles CO) = 986L x (1/2) = ?? L O2.
(NOTE: If reactants and products are in the form of gases and T and P are the same, one may treat volume (liters) as moles.
To answer these questions, we need to use stoichiometry, which is a method for calculating the quantities of reactants and products in a chemical reaction.
First, let's look at the balanced chemical equation for the conversion of carbon monoxide (CO) to carbon dioxide (CO2):
2 CO + O2 → 2 CO2
From the balanced equation, we can see that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.
Now, let's solve the first question:
Q1: How many moles of oxygen gas are needed to convert 50.0 mol of carbon monoxide to carbon dioxide?
Since the stoichiometry of the reaction is 2:1 (2 moles of CO react with 1 mole of O2), we can set up a ratio:
2 mol CO / 1 mol O2 = 50.0 mol CO / x mol O2
By cross-multiplying and solving for x, we find:
x = (50.0 mol CO * 1 mol O2) / 2 mol CO = 25.0 mol O2
Therefore, 25.0 moles of oxygen gas would be needed to convert 50.0 mol of carbon monoxide to carbon dioxide.
Moving on to the second question:
Q2: What volume of oxygen gas would be needed to react with 986 L of carbon monoxide?
To solve this question, we need to use the ideal gas law, which states:
PV = nRT
Where:
P = pressure (assume constant)
V = volume
n = number of moles
R = ideal gas constant
T = temperature (assume constant)
Since the moles of CO and O2 are in a 2:1 ratio based on the balanced equation, we can use the same ratio to find the number of moles of O2 needed:
n(O2) = 1/2 * n(CO) = 1/2 * 986 L CO = 493 L O2
Therefore, the volume of oxygen gas needed to react with 986 L of carbon monoxide is 493 L.
Remember to always balance the chemical equation first and then use stoichiometry or the ideal gas law to solve these types of problems.