When you add 0.320 g of sodium metal to an excess of hydrochloric acid, you find that 3330 J of heat are produced. What is the enthalpy of the reaction as written below?
2Na(s) + 2HCl(aq) --> 2NaCl(aq) +H2(g
As written? As written, is says the heat of reaction is 3330J per .320g Na.
lets do a proportion.
2molesNa/.320g=Hr/3330J
change 2 moles Na to grms, solve for Heat of Reaction.
To determine the enthalpy of the reaction, you need to use the equation for heat transfer:
q = m * c * ΔT
Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, the heat produced (3330 J) is given, but it is important to note that heat is usually reported in joules per mole (J/mol). So, you need to calculate the moles of sodium used in the reaction.
To find the moles of sodium, you can use the molar mass of sodium (22.99 g/mol) and the mass of sodium used (0.320 g):
moles of sodium = mass of sodium / molar mass of sodium
moles of sodium = 0.320 g / 22.99 g/mol
Now that you have the moles of sodium, you can calculate the enthalpy of the reaction:
Enthalpy = heat transferred / moles of sodium
Enthalpy = 3330 J / (0.320 g / 22.99 g/mol)
Simplifying,
Enthalpy = 3330 J * (22.99 g/mol) / 0.320 g
Therefore, the enthalpy of the reaction as written is equal to the calculated value.