When 0.689 g of Ca metal is added to 200.0 mL of 0.500 M HCl(aq), a temperature increase of 112C is observed.
Assume the solution's final volume is 200.0 mL, the density is 1.00 g/mL, and the heat capacity is 4.184 J/gC.
(Note: Pay attention to significant figures. Do not round until the final answer.)
The molar heat of reaction, H rxn, for the reaction of Ca(s) + 2H+(aq) Ca2+(aq) + H2(g) is __________KJ/mol?
q = mass x specific heat x delta T.
You have volume of water and density and that gives you mass H2O, specific heat you have and delta T you have. Calculate q.
Then q/0.689 = joules/g. Multiply that by atomic mass Ca to obtain J/mole and convert to kJ/mole.
your wrong dr bob
To find the molar heat of reaction (Hrxn) for the given reaction, we need to use the equation:
q = m × Cp × ΔT
where:
q is the heat transferred (in Joules),
m is the mass of the reacting substance (in grams),
Cp is the heat capacity of the solution (in J/g°C),
and ΔT is the change in temperature (in °C).
First, let's calculate q using the equation provided:
q = m × Cp × ΔT
= 0.689 g × 4.184 J/g°C × 112°C
Next, we need to convert the amount of Ca from grams to moles. The molar mass of Ca is 40.08 g/mol.
moles of Ca = mass of Ca / molar mass of Ca
= 0.689 g / 40.08 g/mol
Now we can find the molar heat of reaction (Hrxn) using the equation:
Hrxn = q / moles of Ca
Let's substitute the values into the equation:
Hrxn = q / moles of Ca
= (0.689 g × 4.184 J/g°C × 112°C) / (0.689 g / 40.08 g/mol)
Simplifying the expression:
Hrxn = (0.689 g × 4.184 J/g°C × 112°C) × (40.08 g/mol) / 0.689 g
Finally, we can calculate the molar heat of reaction (Hrxn) in KJ/mol:
Hrxn = [(0.689 × 4.184 × 112) × 40.08] / 0.689
= (31.725424 × 40.08) / 0.689
= 1271.67793912 / 0.689
= 1846.564 KJ/mol (rounded to four significant figures)
Therefore, the molar heat of reaction (Hrxn) for the reaction of Ca(s) + 2H+(aq) → Ca2+(aq) + H2(g) is approximately 1846.6 KJ/mol.