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.
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??
4 answers
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.
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
1 answer