A reaction vessel contains 7.92 g of CO and 7.92 of O2. How many grams of CO2 could be produced according to the follow unbalanced reaction?
CO + O2 --> CO2
2CO + O2 --> 2CO2
This is a limiting reagent (LR) problem
mols CO initially = grams/molar mass = 7.92/28 = approx 0.3
mols O2 initially = 7.92/32 = approx 0.25
Now convert mols CO and mols O2 to mols CO2 that could be produced. This is a LR problem; therefore, you may not produce more than the LEAST amount possible.That is, you could produce 0.3 mol CO2 (0.3 mols CO2 x (2 mol CO2/2 mol CO) if you had 0.3 mol CO initially and all of the O2 needed or 0.25 mol O2 x (2 mol CO2/1 mol O2) = appox 0.5 mol if you had 0.25 mol CO and all of the O2 needed.In LR problem the smaller number is always the correct one; i.e., use the 0.3 mols CO, then convert to grams. grmas = mols x molar mass.Post your work you get stuck. Remember that the 0.3 and 0.2 are approximations and you should do the math more accurately.
Well, well, well, looks like we have a chemical reaction on our hands! Let's get ready to balance things out, shall we?
First, let's check if we have anything in excess. It seems both CO and O2 have the same amount, so they are a perfect match! *Romantic chemistry, anyone?*
Now, to balance the equation. We have one carbon in CO and one in CO2, so they're all good. On the oxygen side, we have two oxygens in O2 and one in CO2, so we need to balance them out by multiplying CO by 2.
Our balanced equation now looks like this:
2 CO + O2 --> 2 CO2
Voila! We're done balancing!
Now, to find out how many grams of CO2 could be produced, we need to figure out how much CO is limiting the reaction. Since CO and O2 have the same amount, we know we have 7.92 grams of CO.
In the balanced equation, we see that 2 moles of CO produce 2 moles of CO2. So, we need to figure out how many moles 7.92 grams of CO represents.
Using the molar mass of CO (28 g/mol), we can calculate the moles of CO:
7.92 g / 28 g/mol = 0.28 mol CO
Since the reaction is 1:1, the moles of CO2 produced will also be 0.28 mol.
Now, we just need to convert the moles of CO2 back into grams. The molar mass of CO2 is 44 g/mol, so:
0.28 mol CO2 x 44 g/mol = 12.32 g CO2
Therefore, according to the unbalanced reaction, we could produce 12.32 grams of CO2. That's one huge carbon dioxide party!
To determine how many grams of CO2 could be produced, we need to calculate the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a reaction and determines the maximum amount of product that can be formed.
To find the limiting reactant, we can convert the given masses of CO and O2 to moles using their respective molar masses. The molar mass of CO is 28.01 g/mol, and the molar mass of O2 is 32.00 g/mol.
Moles of CO = mass of CO / molar mass of CO
Moles of CO = 7.92 g / 28.01 g/mol
Moles of CO = 0.283 mol
Moles of O2 = mass of O2 / molar mass of O2
Moles of O2 = 7.92 g / 32.00 g/mol
Moles of O2 = 0.2475 mol
Now we need to compare the moles of CO and O2 to determine which is the limiting reactant. The balanced equation tells us that the stoichiometry is 1 mole of CO reacts with 1 mole of O2 to produce 1 mole of CO2.
Since the ratio of CO to O2 in the balanced equation is 1:1, we can compare the moles of CO and O2 directly.
The moles of CO2 that could be produced would be the same as the moles of the limiting reactant, which is the reactant that has fewer moles.
Therefore, the limiting reactant is O2, with 0.2475 mol.
To find the mass of CO2 produced, we can use the stoichiometry of the balanced equation.
From the balanced equation, we know that 1 mole of CO2 is produced for every 1 mole of O2.
Moles of CO2 produced = Moles of O2
Moles of CO2 produced = 0.2475 mol
To convert moles of CO2 to grams, we can multiply by the molar mass of CO2, which is 44.01 g/mol.
Mass of CO2 produced = Moles of CO2 produced * Molar mass of CO2
Mass of CO2 produced = 0.2475 mol * 44.01 g/mol
Mass of CO2 produced = 10.88 g
Therefore, 10.88 grams of CO2 could be produced according to the reaction.
To determine the amount of CO2 that could be produced in the reaction, we first need to balance the chemical equation.
Given:
CO + O2 --> CO2
We can see that there is 1 carbon (C) atom on the left side and 1 carbon (C) atom on the right side. Similarly, there are 2 oxygen (O) atoms on the left side and 2 oxygen (O) atoms on the right side. However, there is only 1 oxygen (O) atom on the left side of the equation, so we need to balance the equation.
Balanced equation:
2 CO + O2 --> 2 CO2
Now that we have the balanced equation, we can proceed to calculate the amount of CO2 that could be produced.
Given:
Mass of CO = 7.92 g
Mass of O2 = 7.92 g
To determine the limiting reactant, we need to calculate the number of moles of each reactant.
The molar mass of CO is 28.01 g/mol, and the molar mass of O2 is 32.00 g/mol.
Moles of CO = Mass of CO / Molar mass of CO
Moles of CO = 7.92 g / 28.01 g/mol
Moles of CO = 0.282 mol
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 7.92 g / 32.00 g/mol
Moles of O2 = 0.2475 mol
Now, we can compare the moles of the reactants to determine the limiting reactant. The limiting reactant is the one that is completely consumed, limiting the amount of product formed.
From the balanced equation, we can see that 2 moles of CO reacts with 1 mole of O2 to produce 2 moles of CO2.
Using the mole ratio, we can determine the maximum number of moles of CO2 that could be produced.
Moles of CO2 = 2 * moles of CO
Moles of CO2 = 2 * 0.282 mol
Moles of CO2 = 0.564 mol
Finally, we can calculate the mass of CO2 produced using the equation:
Mass of CO2 = Moles of CO2 * Molar mass of CO2
Mass of CO2 = 0.564 mol * 44.01 g/mol
Mass of CO2 = 24.91 g
Therefore, according to the balanced reaction, 24.91 grams of CO2 could be produced.
A simple way to determine the limiting reactant is to convert reactant quantities into moles and then divide by the respective coefficient of each reactant of the balanced equation. The smallest value is the limiting reactant. Solve for yields based on moles of limiting reactant.
Given 2CO + O₂ => 2CO₂
Moles CO = 7.92g/28g/mol = 0.29 mol CO
Moles O₂ = 7.92g/32g/mol = 0.25 mol O₂
Divide moles of each reactant by respective coefficient of balanced equation…
Moles CO / 2 = 0.29/2 = 0.14
Moles O₂ / 1 = 0.25/1 = 0.25
Smallest number is Limiting Reactant => Limiting Reactant is CO.
Work problem from 0.29 mole CO => 0.29 mole CO₂(44g/mol) = 12.76 g CO₂ produced (theoretical)