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 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

$44 g C_3H_8 \times\frac{mol}{44.1gC_3H_8} = 0.998mol C_3H_8$

$0.998mol C_3H_8 \times\frac{5 mol O_2}{1 mol C_3H_8} = 4.987 mol O_2$

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

$4.987 mol O_2 \times\frac{g}{32 mol}= 159.6 g O_2$

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

Answer 2

the answer is 160 grams