Question

I need help with this enthalpy question on Hess' law (in attached photo)

Answer 1

\[Mg+2HCl \rightarrow MgCl2+H2 \]
\[MgO+2HC \rightarrow MgCl2 +H2O \]
\[Q=mc \Delta T\]
\[Q of Reaction 1=100*4.18*(32.2-21.3) \]
\[Q of reaction 2=100*4.18*(28.8-21.4)\]
\[Number of Moles=Mass(g)/Molar Mass(g/mol)\]
\[ number of moles of Mg-0.244/24\]
\[Number of Moles=0.996/(24*1)+(16*1)\]
\[\Delta H=(Q/1000)/Number of Moles\]

Answer 2

Is everything clear till now @hihellohi ? :)

Answer 3

Ok yes thanks again! I just thought that both trials were supposed to be combined. thanks again!

Answer 4

I have one question, I was wondering where 4.18 came from?

Answer 5

they are going to combine that's true, i was just making sure that everything is clear about what i have wrote.
2Mg+O2-->2MgO
MgO+2HCl-->H2O+MgCl2
Mg+2HCl-->MgCl2+H2
2H2+O2-->2H2O
Reverse reaction 1
H2O+MgCl2-->MgO+2HCl
Mg+2HCL-->H2+MgCl2
2H2+O2-->2H2O
Multiply reaction 1 and 2 by 2
2H2O+2MgCl2-->2MgO+4HCL
2Mg+4HCl-->2H2+2MgCl2
2H2+O2-->2H2O
Cancel out like compounds and add Delta H
2Mg+O2-->2MgO

Answer 6

i forgot something when you multiply by 2 the whole equation don't forget to multiply delta H also by 2 and when you reverse equation your reverse the sign.

Answer 7

specific heat capacity of water (4.18 kJ kg-1 °C-1) if c is not provided we use the specific heat capacity of water

Answer 8

thanks a lot! I understand it now!

Answer 9

your most welcome good to hear that :)

Answer 10

May I ask one more question? For my final answer for delta H, I got 659 KJ/mol. Should it be -659 KJ/mol? because I have to do percent error and the accepted theoretical value is -601.6 KJ/mol

Answer 11

to be honest i do know about the last part.

Answer 12

but if the answer is +659 KJ/mol then its +659KJ/mol

Answer 13

Oh ok. I'll retry it then