On a recent holiday I used one kilogram of calor gas (butane) while cooking.?

Question

a) What volume of water was produced?
b) What volume of carbon dioxide was produced?
c) What assumptions have you made in the above calculations?

Answer 1

I'll start with c) The biggest assumption I made is that since it is not my home work, I can do some significant rounding and estimating. Actually one could make the case that "one kilogram" has only one significant figure.
Looking up Butane I find it has a molar mass of 58.1g/mol so we have 17.2 moles of butane, butane has the formula C4H10, so I figure (but I am not a chemist so I may screw the whole thing up at this point) we have 17.2 * 4 = 69 moles of carbon and 17.2 * 5 = 86moles of H2 (molecular hydrogen)

a) 86 moles of H2 would produce 86 moles of H2O with a molar mass of 18g/mol, so we have 1.55kg of water, if condensed to room temperature liquid that would be about 1.5liters
b) 69 moles of carbon would produce 69 moles of CO2 with a molar mass of 44g/mol or about 3kg, at room temperature, CO2 gas has a density of about 1.98kn/m^3, so we have about 1.5m^3 of CO2

to complete the above reactions we would need 86 + 2 * 69 = 224 moles of (atomic) oxygen with a molar mass of about 16g/mol, 224 * 18g = about 3.6kg adding the 1kg of butane, we have 4.6kg, comparing this to the sum of the H2O and CO2 form above we see the effects of rounding, but we are pretty close.

Answer 2

I agree with the previous answers that this should have been asked in Chemistry.

The only thing I would add to the very full answer already given is that you are probably intended to use (as I would have)

PV = nRT

to find the volume of carbon dioxide. I would then be assuming that CO2 behaved near enough like an ideal gas, and I would also have to assume something about temperature and pressure.

Answer 3

In engineering?

it involves a chemical reaction and molecular weights
hardly a suitable answer in engineering