Please reply, this is a part of my Chemisty exam study questions?

Question

Chloride dioxide CLO2 is a reactive oxidizing agent. It is used to purify water.

6CLO2(g) + 3H20(l) = 5HCL03(aq) + HCL(aq)

a) If 71.00g of CLO2 is mixed with 19.00g of water, what is the limiting reactant?
b) What mass of HCLO3 is expected in part (a)?
c) How many molecules of HCL are expected in part (a)?

I don't know how to get the answer, but the end result is posted at the back of the book.
a) CLO2
b)74.1g
c) 1.06 x 10 ^23

I know how to do question one and question 2, thanks Mad Mac, but I still have absolutely on idea on how to get question 3!!!!!!!!!!!!.
Thank you so much to everyone wh0 has helped me so far.

Answer 1

first... look at the mole ratio 6 ClO2 vs. 3 mol water, so u need twice as much ClO2 , the one with the least is "limiting reagent"
a.) 71 g X 1mol/ 67.5 g = 1.052 mol Cl02
vs. 19g X 1 mol/ 18 g = 1.0556 mol H2O
now, given the knowlege that ClO2 consumes Twice as much... u know that 1.052 mol will run out b4 H20 runs out... therefore, ClO2 is the limiting reagent..
b.) mass of HClO3 ( products) ok... use the limiting reagent mole because once it gets consume, the whole reaction stops..
1.052 mol ClO2 X 5 mol HCLO3/ 6 mol ClO2 = .876667 mole HCLO3 ... now, multiply it by the molecular weight ( 83.5) = 74.0783 grams of HCLO3
c.) HCL formation is dictated by how many moles of (limiting reagent formed) relative to its formation... 6 mole of ClO2 mol form = 1 mol of HCl so ... 1.052 mole of ClO2 X 1 mol HCL/6 mole ClO2 = .1753333 mole of HCL <-- moles...
molecules vs. moles = 6.02X10^23 so .1753333 X 6.02 x 10^23 = 1.0555 X 10 ^ 23 molecules of HCL ...

Answer 2

a) The limiting reactant (also called the limiting reagent) is the one that will run out first. Atomic Wts.: Cl = 35.5, O = 16, H = 1.

71.00gClO2 x 1mol ClO2/67.5gClO2 = 1.05mol ClO2

19.00gH2O x 1mol H2O/18gH2O = 1.05

Every 3 mol H2O reacts with 6 mol ClO2 (the balanced equation), so every 1 mol H2O needs 2 mol ClO2. There is 1.05mol H2O. It needs 2.10mol ClO2 to react completely. There is only 1.05mol ClO2. Therefore, ClO2 will run out first and is the limiting reagent.

b) 1.05mol ClO2 x 5mol HClO3/6mol ClO2 x 79.5gHClO3/1mol HClO3 = gHClO3

We already found that ClO2 is the limiting reagent. So the 5/6 factor comes from the balanced equation based on ClO2.

c) 1.05mol ClO2 x 1mol HCl/6mol ClO2 x [6.02x10^23 molecules HCl]/1 mol HCl = molecules HCl

Again the limiting reagent is ClO2. 6.02x10^23 is Avogadro's number. This is the number of molecules in a mole (gram molecular weight) of any substance.

Answer 3

ClO2 is usually the limiting reactant as it is usually used with an excess of water. This is a tricky question.

1. From the equation set up the proportionality:
1/6x(71.00 gr ClO2)/Molecular Wt ClO2 = Moles ClO2 .17544 moles
1/3(19grH2O)/ Molecular Wt H2O = Moles H2O
or, .3330 moles. So ClO2 is limiting Answer

2. 1/5x(gr HClO3)/Molecular Wt HClO3 =.17544 = 5x 84.58 x .17544 = 74.193 gr. Answer

3. There are 6.023 x 10 to the 23 rd power molecules per mole so this times .17544 moles gives total # of molecules of HCl

Answer 4

I know how to do that but I don't feel like doing chemistry right now sorry. If no one gives you a good answer you can e mail me though.