The balanced combustion reaction for C 6 H 6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ If 8.100 g C 6 H 6 is burned and the heat produced from the burning is added to 5691 g of water at 21 ∘ C, what is the final temperature of the water?

The final temperature of water = 35.2 °C

Explanation:

Step 1: Data given

Mass of C6H6 = 8.100 grams

Mass of water = 5691 grams

Temperature = 21 °C

Step 2: The balanced equation

2C6H6(l) + 15O2(g) ⟶ 12 CO2 (g) + 6H2O (l) + 6542 kJ

Step 3:

Q = m*c*ΔT.

⇒with Q = the heat released during this reaction (this depends on the amount of reactants used)

⇒ with m=  the mass of the water

⇒with c = the "specific heat" of water = how much energy it takes to raise the temp of 1g of water by 1°C

⇒with ΔT = the change in the temperature of the water

For every 2 moles of C6H6 consumed, 6542 kJ of heat is released.  

Step 4: Calculate moles for 8.100 grams

8.100grams / 78.11 g/mol= 0.1037 mol es

So, according to the equation, the amount of heat released is:

(0.1037 moles / 2 moles)* (6542 kJ) = 339.2 kJ

Step 5: Calculate the final temperature

Q = mcΔT

ΔT = Q / (m*c)

T2- T1 = Q / (m*c)

T2 = [Q / (m*c)] + T1 = [(339.2 kJ) / (5691g)(0.004186 kJ/g°C)] + 21°C = 35.2°C

The final temperature of water = 35.2 °C

idk i’m struggling too

fe 55.85 g/mol and h2o 18.02 g/mol.