Question

# In a constant-pressure calorimeter, 55.0 mL of 0.330 M Ba(OH)2 was added to 55.0 mL of...

In a constant-pressure calorimeter, 55.0 mL of 0.330 M Ba(OH)2 was added to 55.0 mL of 0.660 M HCl. The reaction caused the temperature of the solution to rise from 23.64 °C to 28.14 °C. If the solution has the same density and specific heat as water (1.00 g/mL and 4.184 J/g·K, respectively), what is ΔH for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Total volume = 55.0 + 55.0 = 110.0 ml

Mass of solution = volume * density

= 110.0 * 1 = 110.0 grams

Q = mass * specific heat * ∆T

= 110.0 * 4.184 * ( 28.14 - 23.64 ) = 2071.08 J

Ba(OH)2 + 2 HCl(aq) = BaCl2(aq) + 2 H2O

Mmoles of Ba(OH)2 = molarity * volume ( in L )

= 0.330 * 55.0 = 18.15

Mmoles of HCl = 0.660 * 55.0 = 36.3

Mmoles of HCl = 2 * mmoles of ba(oh)2 ( according to equation )

So, 18.15 * 2 = 36.3 which is we actually have

Now mmoles of H2O = mmoles of HCl ( look balance equation)

Now, 36.3 mmoles produces = 2071.08 J

1 mole will produce = ( 2071.08 ) / ( 36.3*10^-3) ( mmol to mol )

= 57054.55 j / mol

Or 57.05 KJ/mol

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