During an experiment, a student adds 2.90 g cao to 400.0 ml of 1.500 m hcl . the student observes a temperature increase of 6.00 °c . assuming that the solution's final volume is 400.0 ml , the density is 1.00 g/ml , and the heat capacity is 4.184 j/g⋅°c , calculate the heat of the reaction, δhrxn .

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl  → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

while cooking a piece of 1/2 pound beef steak, a cast iron pan (2.4kg in weight) was heated 260 celsius.