Question 8 is solved, thanks to Eyelessz
I need help with these 2 above questions.
Hah, i might be wrong but heres my take on this question.
Mole Ratio:
2 moles of Na = 1 mole of H(2)
Amt of Na: 0.2 mol
Amt of H(2) = 0.2/2
= 0.1 mol
at r.t.p, 1 mol = 24 dm^3 for gases
vol of hydrogen gas produced : 0.1 x 24 = 2.4 dm^3
Thanks you for the assitance and I have updated this thread with 2 more question which I need help.
qn 11) write out the eqn : H2So4 => 2 H+ + So4(2-)
Mole ratio:
2 H+ = H2So4
Conc. of H+ ions = 2 x 0.05 = 0.1 mol/dm^3
Since they want it in g/dm^3, mutiply by the molecular mass (in this case Hydrogen = 1)
= 0.10 g/dm^3
Then again, i could be wrong lol
O2 + 2CO --> 2CO2
so, between O2 and CO, CO is limiting reagent
20cm3 of CO used up in reaction
20 / 2 = 10cm3 of O2 used up in reaction => 10cm3 of O2 left
Mole ratio of CO and CO2 is 2:2 = 1:1
=> 20cm3 of CO2 produced.
so in the end you got 10cm3 O2, 0cm3 CO and 20cm3 CO2 (C)
For the center question , i'm uncertain. But my answer is C.
Lets see, 20 cm^3 of oxygen translate to be 0.8333 moles (when divided by 24)
20 cm^3 of Co2 translate to be 0.8333 moles too.
If i'm correct, the eqn should be 2 CO + O2 => 2 Co2
Since 1 mole of oxygen requires 2 mole of CO, CO is the limiting reagent. (Since both have 0.8333 mole)
Oxygen will use up half of that 0.8333 mol during reaction, and the other half goes unreacted.
I'm lazy to caculate, since 20cm^3 = 0.8333 mol, half of that remaining = 10 cm^3
Then now we compare mole ratio between CO and CO2
you'll realise its 2 CO = 2 CO2 in the equation
Amt of CO2 = 0.8333 mol
Vol of CO2 formed = 0.8333 x 24 => 20 cm^3
Wait for other forumers to double check lol.
SBS261P's way is shorter, he used mole ratio = volume ratio so i guess his method is preferable to save time during exams. Btw , this O lvl stuff?
Thank you for all the assistance, does anyone have any recommendation
on where can I find reliable tutors to teach me for my O's?
Originally posted by symx:Thank you for all the assistance, does anyone have any recommendation
on where can I find reliable tutors to teach me for my O's?
Oh absolutely. You've asked the right question at the right place.
Here's a reliable, effective, excellent tutor for Chemistry & Biology :
http://infinity.usanethosting.com/Tuition/
Here's a reliable, effective, excellent tutor for Physics & Mathematics :
http://www.freewebs.com/strategictuition/
Do u have any recommedation for tutors in the west area?
Originally posted by symx:Do u have any recommedation for tutors in the west area?
Pity about the location factor. (I'm in the East and Eagle is up North).
Try googling "tuition agency Singapore" and visit the several hundred tuition agencies' websites and in each, narrow down your search by keying the location (west) for the tens of thousands of tutors currently available in Sg.
Thanks for the advice, I will try searching there.
Originally posted by Eyelessz:SBS261P's way is shorter, he used mole ratio = volume ratio so i guess his method is preferable to save time during exams. Btw , this O lvl stuff?
Actually for gas right, it's possible to just use the mole : volume ratio because under standard condition or room temperature and atmospheric pressure, I believe the number of moles for each compound are the same.
Like number of moles for 30cm3 of O2 and always the same as 30cm3 of NO2. This is because the calculation for number of moles of a compound in gaseous state is its volume divided by the molar volume(24dm3) which is the same for both compounds.
Correct me if I'm wrong. Thanks. =)