Chemistry 3/4 2013 Thread

[teletubbies_95]

Oh woah! Dude that's a great question and a great answer !
Thanks !!

[Math Nerd]

How did you find it on google?

[KevinooBz]

How did you find it on google?

I just googled part of the question and there was a link to an atarnotes thread. It must have been a tricky question as it was asked twice.

[Math Nerd]

Ok this question will certainly test you guys.
Some antacids have as their primary active ingredient the compound Na.Al.(OH)2.CO3(s). This compound reacts with stomach acid according to the equation:
NaAL(OH)2CO3(s)+4HCL(aq)->NACl(aq)+AlCL3(aq)+3H2O(l)+CO2(g)

What mass of this compound is required to neturalise 25.0mL of 0.100M HCL(aq)

[Stick]

n(HCl)=c*v=0.1*0.025=0.0025mol
n(NaAl(OH)2CO3)=0.000625mol
m(NaAl(OH)2CO3)=n*M=0.000625*144=0.0900g

[Math Nerd]

How did you get n(NaAl(OH)2CO3)=0.000625mol?

[Stick]

Mole ratio in the equation. 1 mole of NaAl(OH)2CO3 reacts with 4 moles of HCl.

[Math Nerd]

Whoops 

[Stick]

Haha. It's alright. Keep the questions rolling.

[Math Nerd]

The equation for the reaction which occurs between aluminium and hydrogen chloride is :
2Al(s)+6HCl(g)->2AlCl3(g)+3H2(g)
When 2.70g of Al(s) reacts with 3.65g of HCl(g), the maximum mass of H2(g) produced would be?

[Stick]

n(Al)=m/M=2.70/27=0.1mol (0.1/2=0.05)
n(HCl)=m/M=3.65/36.5=0.1mol (0.1/6=0.016667)
Therefore HCl is the limiting reactant.
n(H2)=0.05mol
m(H2)=n*M=0.05*2=0.1g

[Limista]

Is the answer ^ 0.3g?

edit: LOL - made error with limiting reactant (ignore my answer pls)

[Stick]

Is the answer ^ 0.3g?

I think you've used the excess reactant by mistake.

[Math Nerd]

Stick's answer is correct! 

[Math Nerd]

Btw, why is n(H2)=0.05? i thought it was 3/2*0.05? I'm horrible at chemistry, can you explain please?