To solve this problem, we should recall that
the change in enthalpy is calculated by subtracting the total enthalpy of the reactants
from the total enthalpy of the products:
ΔH = Total H of products – Total H of reactants
You did not insert the table in this problem, therefore I
will find other sources to find for the enthalpies of each compound.
ΔHf CO2 (g) = -393.5 kJ/mol
ΔHf CO (g) = -110.5 kJ/mol
ΔHf Fe2O3 (s) = -822.1 kJ/mol
ΔHf Fe(s) = 0.0 kJ/mol
Since the given enthalpies are still in kJ/mol, we have to
multiply that with the number of moles in the formula. Therefore solving for ΔH:
ΔH = [<span>3 mol </span><span>( − </span><span>393.5 </span>kJ/mol<span>) + 1 mol (</span>0.0
kJ/mol)<span>] − [</span><span>3 mol </span><span>( − </span><span>110.5 </span>kJ/mol<span>) + </span><span>2 mol </span><span>( − </span><span>822.1 </span>kJ/mol<span>)]</span>
ΔH = <span>795.2
kJ</span>
Answer: 0.0826mol
PV=nRT
n=PV/RT
n=(1atm)(2.1L)/(310K)(0.082057L*atm/mol*K)=0.0826mol
1. Write out the formula
Pb(NO3)2 (aq) + 2HCl (aq) ----> PbCl2 + 2HNO3
2. Use solubility guidelines (gotta memorize 'em) for the products to see if a solid forms
Nitrates are always soluble so 2HNO3 (aq)
Chlorides (Cl) are always soluble except for when you mix them with copper, lead, mercury, or silver.
Since you mixed it with lead (Pb) it is solid and forms a precipitate. PbCl2 (s)
There is no element in 2s3
