Answer to Question #100215 in General Chemistry for ezgi

Question

The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2(g), and H2O(l):
2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l)
When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter
increases from 25.0°C to 39.5°C. In a separate experiment the heat capacity of the calorimeter is measured to
be 7.794 kJ/°C. Calculate the heat of reaction for the combustion of a mole of CH6N2. (CH6N2: 46.1 g/mol).

Answer 1

Solution.

$Q(cal.) = Q(reac.)$

$Q(cal.) = C(cal.) * \Delta T = 113.013 kJ$

$Q(reac/mol) = \frac{Q(reac.)}{n(CH6N2)}$

$n(CH6N2) = \frac{m(CH6N2)}{M(CH6N2)}$

$n(CH6N2) = \frac{4.00}{46.1} = 0.09 \ mol$

$Q(reac/mol) = \frac{113.013 \times 10^3}{0.09} = 1.30 \times 10^6 \ J/mol$

Answer:

$Q(reac/mol) = 1.30 \times 10^6 \ J/mol$