# Chemistry QuestionWatch

This discussion is closed.
#1
What mass of Aluminium metal will react exactly with 25cm^3 of 0.012 M sulphuric acid? (RAM: Al = 27). Use the equation

2Al(s) + 6H+ (aq) ----> 2Al3+ (aq) + 3H2(g)

0
14 years ago
#2
(Original post by lucaz)
What mass of Aluminium metal will react exactly with 25cm^3 of 0.012 M sulphuric acid? (RAM: Al = 27). Use the equation

2Al(s) + 6H+ (aq) ----> 2Al3+ (aq) + 3H2(g)

n = cV
so if c=0.012 and V=25
then n=0.3

if 2 moles of Al is required to react with 6 moles of the acid then using proportion x=0.6/6 = 0.1 moles
n=m/M so 0.1 = m/27 = 2.7

I think this is it though I haven't done this in a while and can't be sure...
0
14 years ago
#3
(Original post by Natalie Lane)
n = cV
so if c=0.012 and V=25
then n=0.3

if 2 moles of Al is required to react with 6 moles of the acid then using proportion x=0.6/6 = 0.1 moles
n=m/M so 0.1 = m/27 = 2.7

I think this is it though I haven't done this in a while and can't be sure...
Not quite. You have to convert the volume to dm^3 from cm^3

So it would be:

mol = vol * conc
=25/1000 * 0.012
=0.0003

ratio of Al to H is 2/6 = 1/3
so mol of Al = 0.0003/3
mol=0.0001

mass = Mr * conc
mass = 27 * 0.0001
mass = 0.0027 g
0
14 years ago
#4
(Original post by lucaz)
What mass of Aluminium metal will react exactly with 25cm^3 of 0.012 M sulphuric acid? (RAM: Al = 27). Use the equation

2Al(s) + 6H+ (aq) ----> 2Al3+ (aq) + 3H2(g)

you don't have suphuric acid in there as far as I can see. use naming rules to set up the equation...balance it. and then use stoic.

Al has a charge of 3 so it would be

Al+H2SO4-> Al2(SO4)3+H2

balance it....

2Al+3(H2SO4) -> Al2(SO4)3 + 3H2

and then set up a fencepost

25cm^3 H2SO4 X 1 mL H2SO4 X .001L H2SO4 X .012m H2SO4
1cm^3H2SO4_____1mL H2SO4______1L H2SO4

that gives you how many moles you have of sulfuric acid (.0003m H2SO4)

and THEN you do a simple mole to mole ratio

(first find how many grams per mole there is in sulfuric acid) 98.078g
and then the second equation.......

.0003mH2SO4 X 3mAl X 26.982gAl
______________2mH2SO4 -------1mAl

but that seems rather small.
0
14 years ago
#5
(Original post by hihihihi)
Not quite. You have to convert the volume to dm^3 from cm^3

So it would be:

mol = vol * conc
=25/1000 * 0.012
=0.0003

ratio of Al to H is 2/6 = 1/3
so mol of Al = 0.0003/3
mol=0.0001

mass = Mr * conc
mass = 27 * 0.0001
mass = 0.0027 g
I think you balanced wrong....there should only be 3 to 2 not 2 to 6!
0
14 years ago
#6
2Al + 3H2SO4--> Al2(SO4)3 + 3H2 !!!!

ANS= 0.0054g
0
14 years ago
#7
(Original post by vinny2256)
2Al + 3H2SO4--> Al2(SO4)3 + 3H2 !!!!

ANS= 0.0054g
how did you get that?

I'm trying to figure what I did wrong!

we got the same equation!

figured it! acidentally put 3m for Al instead of 2.....and vice versa for sulphuric acid!
0
14 years ago
#8
how did you get that?

I'm trying to figure what I did wrong!

we got the same equation!

figured it! acidentally put 3m for Al instead of 2.....and vice versa for sulphuric acid!
Good!
0
14 years ago
#9
(Original post by vinny2256)
Good!
thank you!

*takes a bow*
0
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