Question
1.20 g of hydrogen gas is completely burned in the presence of excess oxygen gas in a bomb calorimeter producing water. The heat capacity of the calorimeter is 9.43 kJ/°C and the temperature of the calorimeter rose from 25.55 °C to 41.97°C. Calculate the enthalpy change in J and in kJ/mol or water produced.
m=120g
c= 9.43
T1= 25.55 C
T2= 41.97 C
I don't know what to do after this
m=120g
c= 9.43
T1= 25.55 C
T2= 41.97 C
I don't know what to do after this
Answers
2H2 + O2 ==> 2H2O
mols H2 gas initially = 1.20/4 = 0.3 and since the ratio of H2O produced to mols H2 initially is 2/2 we must have produced 0.3 mols H2O.
q = heat capacity x (Tfinal-Tinitial) = ? kJ = dH rxn.
So dH rxn in kJ is for 0.3 mol H2O produced so dH rxn/0.3 gives you dH rxn in kJ/1 mol. Then convert to J also.
Post your work if you get stuck.
mols H2 gas initially = 1.20/4 = 0.3 and since the ratio of H2O produced to mols H2 initially is 2/2 we must have produced 0.3 mols H2O.
q = heat capacity x (Tfinal-Tinitial) = ? kJ = dH rxn.
So dH rxn in kJ is for 0.3 mol H2O produced so dH rxn/0.3 gives you dH rxn in kJ/1 mol. Then convert to J also.
Post your work if you get stuck.
n=m/M = 1.2/4 =0.3 mols
q=ct = (9.43)(16.42)
I am stuck here
q=ct = (9.43)(16.42)
I am stuck here
Related Questions
one moles of a compound containing only carbon, hydrogen, and nitrogen was burned in the presence of...
10cm3 of hydrogen are burned in oxygen to form water.
What volume of oxygen is need to burn hydr...
A 1.2 g sample of C-5 H-10 O is burned completely in excess oxygen to yield CO2(g) and H2O(l) at 25...