easy as pie.. apparently!

ellis182 · Junior Member Joined: 16 Mar 2005 Posts: 4

Hey guys im struggling with these two questions! my textbook is so [poor quality] its from 1982 and [very] useless! can anyone point me in the right direction here?

1) Determine "x" in the following half equation.

x + 14H+ + 6e- 2Cr3+ + 7H2O

2) Chlorine gas can be prepared in the laboratory by the reaction of
manganese dioxide with hydrochloric acid:

MnO2(s) + 4HCl(aq) MnCl2(aq) + 2H2O(l) + Cl2(g)

How much MnO2 should be added to excess HCl to obtain 275 mL of
chlorine gas at 5.0oC and 86.6 kPa?

Any help would be greatly appreciated!

*[i]I moderated your language. Please remember kids have access to this forum. - Rob[/i]

Nick · Regular Joined: 06 Apr 2005 Posts: 23 Location: Spain atm

it would be fine if you used arrows.. --> makes life more simple
i cant read the first equation

for the second

use the formula

p*V = n*R*T V in m3, T in Kelvin, ...
n=p*V/(R*T) then you get the moles of Cl2
if you look at your equation you see how many moles of MnO2 you need for one mole of Cl2....

the rest is up to you,

charco · Regular Joined: 07 Mar 2005 Posts: 96

check out the number of oxygens on the right hand side = 7 so there must be 7 on the left hand side!

check out the number of chromiums (atoms/ions) on the right hand side = 2 therefore there must be two on the lhs

now check out the electrical charges on the right han side compared to the lhs ... you need two negatives on the lhs

hence Cr2O7(2-)

this is the standard reduction of the dichromate ion half equation (orange to green)