Question

A container made from 900g iron is 28.7L in size and has 1.00atm N2 inside. Then CH4 is burned to heat the container and the final pressure is 927 torr and 335K in temp. CH4 + 2 O2 --> CO2 + 2H2O, ΔH_rxn = -890 kJ. S_N2 = 1.04 J/g°C, S_Fe = 0.4498 J/g°C. All heat from burning goes into the container and gas. What was the mass of CH4 in grams required?

a) 225 g
b) 450 g
c) 675 g
d) 900 g

Answer 1

The mass of CH4 required is 16.04 grams.

Step-by-step explanation:

To calculate the mass of CH4 required, we need to use the stoichiometry of the combustion reaction and the enthalpy change of the reaction. From the balanced equation:

CH4 + 2 O2 → CO2 + 2 H2O

we can see that for every mole of CH4 burned, 1 mole of CO2 is produced. Since the enthalpy change of the reaction is -890 kJ, we can use the equation:

ΔH = nΔH_rxn

where ΔH_rxn is the enthalpy change per mole of CH4:

-890 kJ = n * (-890 kJ/mol)

Solving for n:

n = 1 mole

Therefore, the mass of CH4 required is equal to the molar mass of CH4:

mass = n * molar mass = 1 mol * (12.01 g/mol + 4 * 1.008 g/mol)

mass = 16.04 g