Question

Instant hot packs work by crystallizing sodium acetate (NaCH3 COO). The molar enthalpy of crystallization for sodium acetate is (-57.7 kJ/ mol) How many grams of sodium acetate are needed to warm 125.0 mL of water from 21.0°C to 35°C

Answer 1

$\begin{aligned} V_w &= 125.0 mL\\ m_w &= V × d = 125.0 mL × 1 g/mL\\ &= 125g\\ ∆T &= 35°C - 21.0°C = 14°C\\ c &= 4.184 J/g.°C \end{aligned}\\$

$\therefore$ Quantity of Heat (Q) needed is

$\begin{aligned} Q &= mc∆T = 125(4.184)(14)\\ &= 7322J \end{aligned}$

molar enthalpy ($H_c$) of $CH_3COONa$ = -57.7 kJ/mol = -57700 J/mol.

from, Q = n($-H_c$)

$7322 = n(57,700)\\ n = 7322/57700 = 0.127\text{ mol}$

$\begin{aligned} \textsf{mass of } CH_3COONa &= n × \textsf{molar mass}\\ &= 0.127 × 82\\ &= 10.414g \end{aligned}$

$\therefore$ 10.414g of sodium acetate is needed.