3.25 g of C3H8 react with 3.50 g of O2.How many grams of H2O are formed? How much of the excess reactant, in grams, remains after the reaction?

Answer 1

GET A BALANCED EQUATION

First, we want to make a balanced equation of the reaction. We see C₃H₈ reacting with O_2, and this tells us that this is a combustion reaction (organic compound~hydrocarbon reacting with oxygen). The products from a combustion reaction (with a hydrocarbon in particular) will always be CO₂ and H₂O. So we should have this:

C₃H₈ + O₂ ---> CO₂ + H₂O

Now, we need to balance it. To do this, write out each element and count how many are on each side of the equation. After that, we use coefficients to balance them, so both sides have the same number of atoms of each element. Start with carbon, then always leave oxygen last.

C₃H₈ + O₂ ---> CO₂ + H₂O

C: 3 | C: 1

H: 8 | H: 2

O: 2 | O: 3

First carbon

C₃H₈ + O₂ ---> 3 CO₂ + H₂O

C: 3 | C: 3

H: 8 | H: 2

O: 2 | O: 7

Now hydrogen

C₃H₈ + O₂ --->3 CO₂ + 4 H₂O

C: 3 | C: 3

H: 8 | H: 8

O: 2 | O: 10

Finally oxygen

C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

C: 3 | C: 3

H: 8 | H: 8

O: 10 | O: 101

So our balanced equation is: C₃H₈ + 5 O₂ ---> 3 CO₂ + 4 H₂O

FIND OUT HOW MUCH EACH REACTANT PRODUCES OF H₂O

Next, we need to find which of the reactants produce the least about of product. This assumes that we are looking at each reactant individually and assume the other reactant is in excess. The one that produces the least amount out of two will be the limiting reagent/reactant. This will determine how much H₂O is produced. We need to convert the grams to moles by using the molar mass of each reactant, and then we need to use the ratios from the balanced equation to convert these moles into moles of H₂O. Lastly, we need to convert these moles into grams of H₂O. This will require some dimensional analysis to help us out.

C₃H₈: 3.25 g x (1 moles/44.1 g/mol) x (4 moles of H₂O/ 1 mole of C₃H₈) x (18.02 g/ 1 mole) = 5.31 g of H₂O

O₂: 3.50 g x (1 moles/32 g/mol) x (4 moles of H₂O/ 5 mole of O₂) x (18.02 g/ 1 mole)= 1.58 g of H2O

So since the O₂ produced the least amount of H₂O, it is our limiting reactant, and the amount of H₂O we expect to produce with these amounts of reactants is 1.58 g.

Lastly, we need to find the amount of C₃H₈ that will remain from this reaction. To do this, we will convert the extra H₂O produced by C₃H₈ compared to O2 into C₃H₈.

5.31g-1.58g= 3.73 g of H2O produced by C₃H₈

3.73 g of extra H₂O x (1 moles/18.02 g/mol) x (1 moles of C₃H₈ / 4 mole of H₂O) x (44.1 g of C₃H₈ / 1 mole)= 2.28 g of C₃H₈ remains

I hope this was helpful!