Chemistry, asked by benjaminsamraj672, 7 months ago

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?​

Answers

Answered by sknirwal
1

Answer:

my friend your answer is given below please mark as brinliest>>>please

Explanation:

X+2Y→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>23.9721.3+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+x21.3+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= 1.86x=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= 19.44m(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}  

16

19.44

​  

=  

z

4.53

​  

 

z= 3.73z=3.73

Answer:3.73 g of Y

please mark as brainliest hope this help you

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