Suppose that coal of density 1.5 g/cm^3 is pure carbon. (it is, in fact, much more complicated, but this is a reasonable first approximation.) the combustion of carbon is described by the equation:
c(s) + o2(g) > co2(g) ∆h° = −394 kj
what is the value of q (heat) when a lump of coal of size 5.6 cm x 5.1 cm x 4.6 cm is burned?
answer in units of kj.

q = -6464.9 kJ

Explanation:

We are given that the heat of combustion is  ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.

vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³

m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g

mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol

q =  −394 kJ /mol C x 16.41 mol C = -6464.9 kJ

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.