When you add 0.567 g of sodium metal to an excess of hydrochloric acid, you find that 5890 J of heat are produced. What is the enthalpy of the reaction as written below?
2Na(s)+2HCl(aq)=2NaCl(aq)+H2(g)
figure the moles of sodium you used.
then
MolesNa/2 liberated 5890J
Hr= -5890J*2/molesYouUsed
5890 J x (2*23/0.567) = ?
477848.3?
That's ok but you have too many significant figures in the answer.
To determine the enthalpy of the reaction, we need to use the equation:
ΔH = q / n
Where:
ΔH = Enthalpy change
q = Heat produced/absorbed
n = Number of moles of reactant or product involved in the reaction
First, we need to determine the number of moles of sodium metal (Na) used in the reaction. We can do this by dividing the mass of sodium (0.567 g) by its molar mass. The molar mass of sodium (Na) is 22.99 g/mol.
Number of moles of Na = mass of Na / molar mass of Na
Number of moles of Na = 0.567 g / 22.99 g/mol
Next, we need to calculate the heat produced (q) in Joules. The given value is 5890 J.
Now, we can substitute the values into the equation to find the enthalpy change (ΔH).
ΔH = q / n
ΔH = 5890 J / (moles of Na from above)
By performing the calculations, you can determine the enthalpy of the reaction as written in the equation.