Question

A 25.00 g solid sample of Ca(OH)2 was added into 1250 mL of 0.400 M HCl aqueous solution. The temperature of the solution was decreased from 36.6oC to 20.0oC. Assuming the final volume of the solution is 1250 mL / the density of the solution is 1.000 g/mL / the heat capacity of the solution is 4.18 J/(oC*g). Please determine the molar heat of the reaction of the limiting reactant.

Answer 1

First write the balance reaction between Ca(OH)₂ and HCl.

Ca(OH)₂ (s) + 2 HCl (aq) $\rightarrow$ CaCl₂ (aq) + 2H₂O (l)

Now first find the moles for Ca(OH)₂ and HCl.

Moles of Ca(OH)₂ = 25.00 g Ca(OH)₂ * (1 mol Ca(OH)₂ / 74.0926 g) = 0.3374 moles

Moles of HCl = Molarity * volume in liter = 0.4 moles / L * 1.25 L = 0.50 moles

Now find limiting reactant by using mole to mole ratio for both reactant.

Since mole to mole ratio for Ca(OH)₂ to CaCl₂ is 1:1. If we use 0.34 moles we get same moles i,e 0.34 moles of CaCl₂ .

Mole to mole ratio for HCl to CaCl₂ is 1:2 . If we use 0.5 moles of HCl then moles of CaCl₂ produced will be,

0.5 mol HCl * ( 1 mol CaCl₂ / 2 mol HCl) = 0.25 mol CaCl₂ .

From above calculation, HCl is limiting reactant due to formation of less moles of products.

To find the molar heat, first find the heat produced by solution.

Since q = m * c * $\Delta T$

First find mass and $\Delta T$ in for given reaction.

Since density = mass / volume

1.00 g / ml = mass / 1250 ml

mass = 1.00 g /ml * 1250 ml = 1250 g

And $\Delta T$ = (T_final - T_initial) Since T(final) = 20 ^oC and T(initial) = 36.6 ^oC

Therefore $\Delta T$ = ( 20 - 36.6) = -16.6 ^oC.

Therefore q = 1250 g * 4.18 J/g ^oC * (-16.6 ^oC) = -86735 J.

Since heat of reaction = q / n

where n - Number of moles

Therefore heat of reaction = -86735 J / 0.50 mol = -173470 J / mol = -173470 J / mol * (1kJ / 1000 J) = -173.470 kJ / mol

After rounding off, final answer will be -174 kJ / mol or -173.5 kJ / mol.

If you find any mistake mention in the comment box.

Thanks.