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Consider the reactions $$X + 2Y \rightarrow 3Z$$$$2X + Z \rightarrow Y.$$ Let $n_{X}$, $n_{Y}$, $n_{Z}$ denote the numbers of molecules of chemicals $X, Y, Z$ in the reaction chamber. Then which of the following is conserved by both reactions?

- $n_{X} + n_{Y} + n_{Z}$.
- $n_{X}+ 7n_{Y} + 5n_{Z}$.
- $2n_{X} + 9n_{Y} − 3n_{Z}$.
- $3n_{X} − 3n_{Y} + 13n_{Z}$.
- None of the above.

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Best answer

Basically here we need to find the number of molecules are same before and after the reaction ie. the conservation of mass.

To check that just take options and eliminate one by one. Following is the breakdown of options

In first reaction put the values of option (a) $n+2n = 3n$ is true for first equation but if we put the values for second equation, we get $2n+n = n$ which is not possible. So, option (a) is conserving mass for equation one but not for equation two.

Similarly, eliminating all the options we can conclude option (b) is correct. As for equation one, $n + 2*7n = 3*5n$, for equation two, $2n + 5n =7n.$ So, mass is conserved for both reactions.

Correct option: B

To check that just take options and eliminate one by one. Following is the breakdown of options

In first reaction put the values of option (a) $n+2n = 3n$ is true for first equation but if we put the values for second equation, we get $2n+n = n$ which is not possible. So, option (a) is conserving mass for equation one but not for equation two.

Similarly, eliminating all the options we can conclude option (b) is correct. As for equation one, $n + 2*7n = 3*5n$, for equation two, $2n + 5n =7n.$ So, mass is conserved for both reactions.

Correct option: B

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