Chlorine dioxide, ClO2, has been used as a disinfectant in air conditioning systems. It reacts with water according to the following equation.
6ClO2 + 3H2O --> 5HClO3 + HCl
If 142.0 g of ClO2 is mixed with 33.0 g of H2O, how many grams of the reactant will remain if the reaction is complete?
do i have to use the masses give and use it with the two products??
This is a limiting reagent problem.
1. Convert 142 g ClO2 to moles.
2. Convert 33.0 g H2O to moles.
3a. Using the coefficients in the equation, convert moles ClO2 to moles HCl (you may use HClO3 if desired but it's simpler to use HCl).
3b. Convert moles H2O to moles HCl the same way.
3c. The smaller number of moles HCl is the correct answer and the value of the reactant that produced that number is the limiting reagent.
4. Using the limiting reagent, convert moles of that to moles of the other reactant (that is, either H2O or ClO2 depending upon the limiting reagent), subtract that from the moles of both ClO2 and H2O that you started with, and convert both to grams. Post your work if you get stuck.
okay i got my answer of 1.75 mol and 3.05 mol but i don't understand part 4
I don't get those answers.
I have molar mass ClO2 = 67.45 g/mole
molar mass H2O = 18.015 g/mol.
mols ClO2 = 142 x (1 mol/67.45) = 2.10 mols ClO2.
mols H2O = 33.0 x (1 mol/18.015) = 1.83 mols H2O.
mols HCl formed if all of the ClO2 reacted is
2.10 x (1 mol HCl/6 mols ClO2) = 2.10 x (1/6) = 0.350 mols HCl.
mols HCl formed if all of the water reacted is
1.83 x (1 mol HCl/3 mols H2O) = 1.83 x (1/3) = 0.61 mols HCl.
The smaller number is 0.350; therefore, ClO2 is the limiting reagent. Thus, all of the ClO2 will be used and zero will remain after the reaction. How much of the water will be used?
That will be mols ClO2 x (3 mols H2O/6 mols ClO2) = 0.350 x (3/6) = 0.175 mols H2O used. We had 1.83 mols H2O to begin; therefore, at the end of the reaction we will have 1.83 -0.175 = ?? mols H2O remain unreacted. Change that to grams by g = mols x molar mass.
3.17 g of H2O remains
Yes, in order to determine how much of the reactant will remain, you need to consider the masses of both ClO2 and H2O and compare them to the stoichiometry of the reaction equation.
First, let's calculate the molar masses of ClO2 and H2O.
- The molar mass of ClO2 is 67.45 g/mol (from the atomic masses of Cl and O).
- The molar mass of H2O is 18.02 g/mol (from the atomic masses of H and O).
Next, we need to determine the moles of ClO2 and H2O given the masses provided.
- Moles of ClO2 = mass of ClO2 / molar mass of ClO2 = 142.0 g / 67.45 g/mol
- Moles of H2O = mass of H2O / molar mass of H2O = 33.0 g / 18.02 g/mol
Now we can use the balanced equation to determine the stoichiometry of the reaction.
According to the equation, 6 moles of ClO2 react with 3 moles of H2O.
Since the ratio is 6:3, we know that for every 6 moles of ClO2, we need 3 moles of H2O. However, in this case, we have an excess of H2O compared to the ratio specified by the equation.
We can determine the limiting reactant by comparing the moles of ClO2 and H2O:
- Moles of ClO2: (142.0 g / 67.45 g/mol) = 2.107 moles
- Moles of H2O: (33.0 g / 18.02 g/mol) = 1.833 moles
Since we have fewer moles of H2O, it is the limiting reactant. This means that all of the H2O will be completely consumed, and there will be excess ClO2 remaining.
To calculate the grams of reactant that will remain, we need to find the amount of ClO2 that reacted with the H2O.
Given the stoichiometry of the reaction, for every 3 moles of H2O reacted, 6 moles of ClO2 will be consumed.
Since we have 1.833 moles of H2O:
- Moles of ClO2 Reacted = (1.833 moles H2O) * (6 moles ClO2 / 3 moles H2O) = 3.667 moles ClO2
Finally, we can calculate the mass of ClO2 that will remain:
- Mass of ClO2 remaining = (moles of ClO2 - moles of ClO2 reacted) * molar mass of ClO2
= (2.107 moles - 3.667 moles) * 67.45 g/mol
Therefore, the mass of ClO2 that will remain if the reaction is complete is the calculated value above.