Question

in an experiment, 44 g of propane, C3Hg, was burned producing 132 g of carbon dioxide and 72g of water. the mass of oxygen that was needed for the reaction was?

Answer 1

the answer is 160 grams

Answer 2

The law of conservation of mass states that mass can neither be created or destroyed, hence we expect product mass to equal reactant mass.

mass of propane + mass of oxygen = mass of carbon-dioxide + water

`44+x= 132+ 72\\\\=> x= (132+72)-44\\=>x= 160g`

Using stochiometry, we first write a balanced the equation

`C_3H_8 + 5O_2 ==> 3CO_2 + 4H_2O`

The ratio of propane to oxygen is 1:5, 1 mol propane reacts with 5 mol oxygen

First find the moles of propane using the mass of propane 44g and molecular mass of propane 44.1g/mol

`44 g C_3H_8 *(mol)/(44.1gC_3H_8) = 0.998mol C_3H_8`

`0.998mol C_3H_8 *(5 mol O_2)/(1 mol C_3H_8) = 4.987 mol O_2`

The grams of oxygen would be, using
`O_2` molecular mass 32 g/mol

`4.987 mol O_2 *(g)/(32 mol)= 159.6 g O_2`

The answer rounded to 3 significant figures gives us
`160g O_2`.