Chemistry 3/4 2013 Thread

[zvezda]

ah yeh of course, that's right.

[Stick]

Keep the questions coming!

[saba.ay]

It's 16 mg , I multiplied on my calc wrong it's 1. 6 x 100 /10= 16mg.

Got it. Had a friend explain it to me. Thanks again.

[teletubbies_95]

Thanks Stick , Starfish and Kevin for your input and sos for the confusion ! I was staring at the question for a full 5 minutes when I first saw it. But now that you guys have broken it down , it really helped. Thanks! and good luck for your SAC Stick , I'm sure you'll do amazingly !

[Stick]

I hope so. Keep posting up your gravimetric analysis problems and I'll try to help you out.

[Math Nerd]

Alkanes are hydrocarbons with the general formula CnH2n+2. An alkane consists of 81.8% carbon by mass, the value of n would be? (math question i made up teehee)

[Stick]

I'm doing this one on my iPad haha.

C:H
=(81.8/12):(18.2/1)
=6.8167:18.2
=1:(8/3)
=3:8

Therefore n=3.

[Math Nerd]

5.00g of a mixture of magnesium and zinc was allowed to completely react with dilute sulfuric acid according to the equation: M(s)+H2SO4(aq)->MSO4(aq)+H2(g)
If 0.250g of hydrogen gas was produced in the reaction, what would the percentage composition of magnesium in the mixture be?

[Jaswinder]

i have tried it, but im probably wrong hehe i got 60.75%

[teletubbies_95]

Yep. I tried. I think it's 60.75%
So what you do is :
1. find the moles of h2
2. do mole ratio with mg
3.use n/mr to find mass of mg
4. then find the percentage

Thanks for asking this question !!!

[KevinooBz]

5.00g of a mixture of magnesium and zinc was allowed to completely react with dilute sulfuric acid according to the equation: M(s)+H2SO4(aq)->MSO4(aq)+H2(g)
If 0.250g of hydrogen gas was produced in the reaction, what would the percentage composition of magnesium in the mixture be?

First convert the mass of hydrogen into mol. 0.250g/2gmol^-1=0.125mol. Next calculate the mol of magnesium, from the given equation it should be 0.125mol as the magnesium and hydrogen react in a 1:1 ratio. Find the mass of  magnesium which is 0.125mol x 24.3gmol^-1=3.0375g. Last step is to find the percentage of mixture which is 3.0375g/5g x 100=60.75%.

[Math Nerd]

37.6% is the correct answer.

[KevinooBz]

37.6% is the correct answer.

Oh lol. So the equation has M not Mg?

[Math Nerd]

yup

[KevinooBz]

This question is quite mathematically involved, but not impossible.

n(H2) = 0.250 / 2 = 0.125 mol

Since Mg and Zn both react on a 1:1 ratio with H2SO4 to release 1:1 H2 gas, this implies

n(Mg) + n(Zn) = 0.125 mol  ----------[1]

Also, the mass of Mg and Zn add up to 5.00g

n(Mg) * 24.3 + n(Zn) * 65.4 = 5.00  ------------[2]

We see that's a set of simultaneous equation, multiplying [1] by 24.3 and eliminate

n(Zn) * 41.1 = 1.9625
n(Zn) = 0.04775 mol
m(Zn) = 0.04775 * 65.4 = 3.12g
%(Zn) = 3.12/5.00 * 100% = 62.4%

%(Mg) = 100% - 62.4% = 37.6%
:. Option C

[was this from TSFX?]

props to Mao for this one, apparently google can find any and everything.