Household bleach contains (NaClO)2 that decomposes in water,
NaClO + h2o = Na+ + OH- + Cl2 + O2
If 15ml bleach containing 5.85℅ NaClO is allowed to decomposes completely, then many mL of o2 at 22C and 755mmHg can be liberated. Density of bleach= 1.12g/ml
Answer
I found the volume of NaClo by multiplying 15ml by 5.45/100
Where do i go from here???
first balance
2 NaClO + H2O --> 2 Na+ +2OH- + Cl2 + O2
How many mols of NaClO ?
Na = 23 g/mol
Cl = 35.5
O = +16
so
NaClO = 74.5 g/mol
2 NaClO = 149 g
then
grams of NaClO = .0585 *1.12*15
= .983 grams of NaCl
.983/74.5 = .0132 mols of NaClO
we get half as many mols of O2
so
0.0066 mols of O2
use the gas law but approximately
22.4 liters/mol * .0066 mol = .148 liter
or 148 mL
I had done the same process previously but my teacher's answer is 74.9mL...do you know any other way to work it out?
Also, when i inputed .0066mol O2 into pv=nrt to get volume, i ended up witj 161.5L. I don't know what im/doing wrong...
Damon didn't quite balance the equation. There are 3 O atoms on the left and four on the right. What was your balanced equation? I'll look at this later.
I re balanced but still not getting the right answer
2naclo + h2o = 2 Na + 2oh + cl2+ 1/2o2
My balanced equation is
4NaClO + 2H2O ==> 4NaOH + 2Cl2 + O2 which is just twice your balanced equation but the ratios are the same.
1.12 g/mL x 15 mL x 0.0585 = 0.982 g NaClO
mols = 0.982/74.45 = 0.0132 mols NaClO
mols O2 = 0.0132/4 = 0.00330
Then V = nRT/P. I used P = 755/760
R = 0.08206
T = 273 + 22 = 295
All of that put together is
V in L = nRT/PV
L = 0.00330*0.08206*295*760/755 = 0.0804 L = 80.4 mL.
I don't believe 74.9 mL is right and note that if I estimate as Damon did that 0.0033 x 22.4 = 73.9 mL.
The only thing that might make a difference is that your teacher is not using the same equation; ie. your teacher's balanced equation is not the same so the ratio's aren't the same but the difference in answers would need to be a whole number or fraction like 2 or 1/2 or 1.5 and the difference we're seeing is just a fraction of that.
To find the volume of O2 liberated, we need to use the stoichiometry of the reaction.
From the balanced equation, we see that 1 mole of NaClO gives 1 mole of O2. We need to find the amount of NaClO in moles first.
To find moles of NaClO:
Moles = Mass / Molar Mass
Given:
Volume of bleach = 15 ml
Density of bleach = 1.12 g/ml
Percent composition of NaClO = 5.85%
First, we can calculate the mass of bleach:
Mass of bleach = Volume of bleach x Density of bleach
Mass of bleach = 15 ml x 1.12 g/ml
Next, we can find the mass of NaClO using the percent composition:
Mass of NaClO = Mass of bleach x (% NaClO / 100)
Mass of NaClO = (15 ml x 1.12 g/ml) x (5.85 / 100)
Now, we can find the moles of NaClO using the molar mass of NaClO:
Molar mass of NaClO = (23 + 35.5 + 16) g/mol
Moles of NaClO = Mass of NaClO / Molar mass of NaClO
Once we have the moles of NaClO, we know that it will produce an equal number of moles of O2. Then we can use the ideal gas law to calculate the volume of O2 in liters:
PV = nRT
Given:
Temperature (T) = 22°C = 22 + 273 = 295 K
Pressure (P) = 755 mmHg
R = 0.0821 L·atm/(mol·K) (gas constant)
Rearranging the equation:
V = nRT / P
Substitute the values in and solve for V.
Finally, convert the volume from liters to mL by multiplying by 1000.
Note: Make sure to convert the units used in the calculations to be consistent (e.g., grams, moles, atmosphere).
Now you have all the steps to solve the problem. Plug in the values you have obtained in each step and perform the necessary calculations to find the volume of O2 liberated.