Chem Question: Mole Concept

symx SG

Question 8 is solved, thanks to Eyelessz

I need help with these 2 above questions.

Eyelessz SG

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

xstryker SG

symx SG

Thanks you for the assitance and I have updated this thread with 2 more question which I need help.

Eyelessz SG

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

SBS261P SG

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)

Eyelessz SG

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.

Eyelessz SG

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?

UltimaOnline SG

Eyelessz and SBS261P are both correct, of course.

And these questions are of easy-to-average difficulty for 'O' levels.

Students expecting to score distinctions and wishing to challenge themselves are invited to attempt the questions (eg. indicated as 'O' level) in my thread :

http://www.sgforums.com/forums/2297/topics/320107

symx SG

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?

UltimaOnline SG

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/

symx SG

Do u have any recommedation for tutors in the west area?

UltimaOnline SG

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.

symx SG

Thanks for the advice, I will try searching there.

d3sT1nY SG

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. =)

SBS261P SG

correct