How many grams of water are produced by the combustion of 4.00 kg of propane

Answer 1

The usual method requires you to balance the equation for full combustion of propane:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

and do the stoichiometry calculation

4000 g P(1 mole/20.09g P)(4 moles W/1 mole P)(18.02 g/mole W) = grams of water

The quicker way is to use the fact that all the grams of H go to H₂O. This approach doesn't need the equation only the mass ratios within the compounds:

4kg (8.08 kg H/20.09 kg P)(18.02 kg W/2.02 kg H) = kg W (obvious here that any mass units can be used. It is also true for the stoichiometric calculation, but the units are problematic and that defeats the purpose of what is generally a simple and error-free method.)

Please consider a tutor. Take care.

Answer 2

First, look up propane to find its molecular formula. It is C₃H₈. Then write a an unbalanced chemical reaction equation using propane. Combustion takes a hydrocarbon plus oxygen (O₂) and yields carbon dioxide (CO₂) and water (H₂O):

C₃H₈ + O₂ → CO₂ + H₂O

Then balance the equation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

Calculate the molar masses of the compounds in question:

C₃H₈ = 3(12.011) + 8(1.008) = 44.097 g/mol

H₂O = 2(1.008) + 1(15.999) = 18.015 g/mol

Calculate the number of moles you have of propane:

4.00 kg = (4000 g) x (1 mol)/(44.097 g) = 90.709 moles

Use the molar ratios in the balanced chemical reaction equation to calculate the number of moles of water.

The reaction equation says for every 1 mole of propane, 4 moles of water are produced. So if you start with 90.709 moles of propane, you will end with (90.709)(4) = 362.84 moles of water.

Convert the 362.84 moles of water into equivalent mass:

(362.84 moles of water) x (18.015 g)/(1 mole) = 6536.5 grams of water

Rounding to 3 sig figs, we get 6540 grams of water