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stellarik [79]
3 years ago
11

The balanced combustion reaction for C 6 H 6 C6H6 is 2 C 6 H 6 ( l ) + 15 O 2 ( g ) ⟶ 12 CO 2 ( g ) + 6 H 2 O ( l ) + 6542 kJ 2C

6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 5.500 g C 6 H 6 5.500 g C6H6 is burned and the heat produced from the burning is added to 5691 g 5691 g of water at 21 ∘ 21 ∘ C, what is the final temperature of the water?
Chemistry
1 answer:
Lana71 [14]3 years ago
4 0

Explanation:

First, we will calculate the molar mass of C_{6}H_{6} as follows.

Molar mass of C_{6}H_{6} = 6 \times 12 + 6 \times 1

                                   = 78 g/mol

So, when 2 mol of C_{2}H{6} burns, then heat produced = 6542 KJ

Hence, this means that 2 molecules of C_{6}H{6} are equal to 78 \times 2 = 156 g of C_{6}H_{6} burns, heat produced = 6542 KJ

Therefore, heat produced by burning 5.5 g of C_{6}H{6} =                  

       6542 kJ \times \frac{5.5 g}{156 g}

            = 228.97 kJ

            = 228970 J           (as 1 kJ = 1000 J)

It if given that for water, m = 5691 g

And, we know that specific heat capacity of water is 4.186 J/g^{o}C .

As,             Q = m \times C \times (T_{f} - T_{i})

          228970 J = 5691 g \times 4.184 J/g^{o}C \times (T_{f} - 21)
^{o}C

                T_{f} - 21^{o}C = 9.616^{o}C

                T_{f} = 30.6^{o}C

Thus, we can conclude that the final temperature of the water is 30.6^{o}C.

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Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

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