Answer :
When 145.0 g of propane (c3h8, 44.1 g/mol) is reacted with 98.5 g oxygen (32.0 g/mol), the theoretical yield, in grams, of water (18.0 /mol), according to the balanced equation below, is 44.28 g.
The balanced chemical equation for the reaction between propane and oxygen to yield water is: C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)The stoichiometric ratio of propane to water is 1:4, meaning that for every 1 mole of propane, 4 moles of water are produced.
Molar mass of propane = 44.1 g/mol
Molar mass of oxygen = 32.0 g/mol
Molar mass of water = 18.0 g/mol
Now let's determine the limiting reactant. To do this, we need to convert the given masses of reactants to moles:
moles of propane = 145.0 g / 44.1 g/mol = 3.29 mol
moles of oxygen = 98.5 g / 32.0 g/mol = 3.08 mol
Since the stoichiometric ratio of propane to oxygen is 1:5, we see that oxygen is the limiting reactant.
Therefore, we will use the moles of oxygen to calculate the theoretical yield of water.
The stoichiometric ratio of oxygen to water is 5:4, meaning that for every 5 moles of oxygen, 4 moles of water are produced.
moles of water = (3.08 mol O2) x (4 mol H2O / 5 mol O2) = 2.46 mol H2O
Molar mass of water = 18.0 g/mol
The theoretical yield of water is therefore:
mass of water = (2.46 mol) x (18.0 g/mol) = 44.28 g
Therefore, the theoretical yield of water is 44.28 g.
For more such questions on propane, click on:
https://brainly.com/question/28358864
#SPJ11
