Asked by Nancy
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 hoe 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???
NaClO + h2o = Na+ + OH- + Cl2 + O2
If 15ml bleach containing 5.85℅ NaClO is allowed to decomposes completely, then hoe 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???
Answers
Answered by
Damon
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
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
Answered by
Nancy
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?
Answered by
Nancy
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...
Answered by
DrBob222
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.
Answered by
Nancy
I re balanced but still not getting the right answer
2naclo + h2o = 2 Na + 2oh + cl2+ 1/2o2
2naclo + h2o = 2 Na + 2oh + cl2+ 1/2o2
Answered by
DrBob222
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.
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.
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