Having trouble understanding this.

Question

Instant hot packs work by crystallizing sodium acetate (NaCH3COO). The molar enthalpy of crystallization for sodium acetate is -56.7 kJ/mol. How many grams of sodium acetate are needed to warm 125.0 mL of water from 21.0oC to 35.4oC?

Answer 1

First determine the mass of the water:

125mL = 0.125L

The density of water at approx 25°C is 0.997 g/mL = 0.997 kg/L

m = ρV = (0.997)(0.125) = 0.124625 kg

Then use:

Q = mC_pΔT where Q is the heat required to heat the water, m is the mass of the water, Cp is the Specific Heat (look it up in a table to find it is 4.184 kJ/kg°C), and ΔT is the change in temperature

Q = (0.124625)(4.184)(35.4 - 21.0) = 7.5086 kJ is required to heat the water

Then switch over to the hot packs.

The molar mass of NaCH₃COO is:

Na = 22.990 g/mol

C = 2(12.011) = 24.022 g/mol

H = 3(1.008) = 3.024 g/mol

O = 2(15.999) = 31.998 g/mol

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Total = 82.034 g/mol

We are told that the molar enthalpy of the crystals is -56.7 kJ/mol. We can ignore the minus sign (just says the it gives off heat). We need to get 7.5086 kJ to heat the water:

(7.5086 kJ)/56.7 kJ/mol) = 0.132427 moles

Since the molar mass is 82.034 g/mol then the mass of crystals we need is (82.034)(0.132427) = 10.9 g