Home > Chemistry > Enthalpy

When you add 0.857 g of sodium metal to an excess of hydrochloric acid, you find that 8910 J of heat are produced. What is the enthalpy of the reaction?

Answer 1

Avery Adams

#DeltaH_"rxn"^@=-238.9*kJ*mol^-1#

We scrutinize the response.

#Na(s) + HCl(aq) rarr NaCl(aq) + 1/2H_2 + Delta#

"Moles of sodium" are calculated as (0.857g)/(22.99gmol^-1) = 0.0373mol.

And thus #DeltaH_"rxn"^@=(8910*J)/(0.0373*mol)=-238.9*kJ*mol^-1#, and we mean per mole of reaction as written.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Avery Adams

The enthalpy of the reaction can be calculated using the heat produced and the amount of substance reacted. In this case, the enthalpy of the reaction can be determined using the equation:

ΔH = q / n

Where: ΔH = enthalpy of the reaction q = heat produced (8910 J) n = number of moles of substance reacted

First, we need to find the number of moles of sodium reacted using its molar mass:

molar mass of sodium (Na) = 22.99 g/mol

moles of Na = mass / molar mass = 0.857 g / 22.99 g/mol

Now, we can calculate the enthalpy of the reaction:

ΔH = 8910 J / moles of Na

Calculate moles of Na first, then substitute into the equation to find ΔH.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.