Answer to Question #90851 in Inorganic Chemistry for Agou

Question

Two pure gases of elemental composition X and Y are allowed to mix in a
chamber. When combined, these two gases produce a liquid with the
chemical formula XY2. Interestingly, the liquid product has no discernible
smell, but the two precursor gases do! Gas X smells like flowers, while gas Y
smells like fish. In one experiment, 21.3 grams of X is mixed with 4.53
grams of Y and 23.97 grams of the product is obtained. The headspace, the
gas above the liquid in the chamber, smelled like roses. How much of gas Y
will be needed for the exclusive formation liquid XY2 (no smells in the
headspace) if the experiment starts with 16g of gas X?

Answer 1

$$X+2Y\rightarrow XY_2$$

According to the law of conservation of mass, in any chemical reaction, the mass of reacting substances at the start of reaction will be the same as the mass of the products at the end.

We can see that

$$21.3+4.53>23.97$$

Then one of the reactants is in excess. As this reactant has the smell of rose and it is left over the liquid, then the reactant in excess is X.

Then the law of conservation of mass will be written as:

$$21.3+4.53= 23.97+x$$

where x- is mass of gas X left after the reaction is completed.

$$x= 21.3+4.53-23.97= 1.86$$

Find mass of gas X that is required to react with 4.53 g of X:

$$m(X)= 21.3-1.86= 19.44$$

So, we need 19.44 g of X to completely react with 4.53 g of Y.

And we have 16 g of X to completely react with z g of Y.

Solve the proportion:

$$\frac{19.44}{16} =\frac{4.53}{z}$$

$$z= 3.73$$

Answer:3.73 g of Y