Question

We know that balancing of a chemical equation is entirely based on law of conservation of mass. However the concept of Principle of Atom Conservation (POAC) can also be related to law of conservation of mass in a chemical reaction. So, POAC can also act as a technique for balancing a chemical equation. For example, for a reaction : $AB{C}_3\to AB+C_2$ On applying POAC for A, B & C and relating the 3 equations, we get : $\frac{n_{A}B{C}_3}{2}=\frac{n_{AB}}{2}=\frac{n_{C_2}}{3}$ $n_x:$ number of moles of X) Thus, the coefficients of $AB{C}_3,AB\&C_2$ in the balanced chemical equation will be 2, 2 & 3 respectively and the balanced chemical equation can be represented as : $2AB{C}_3\to 2AB+3C_2$ Now answer the following question :

If the atomic masses of X and Y are 10 & 30 respectively, then the mass of $X{Y}_3$ formed when 120 g $Y_2$ reacts completely with sufficient amount of X is : Reaction$X+Y_2\to X{Y}_3$

• A. 133.3 g
• B. 200 g
• C. 266.6 g
• D. 400 g

Answer 1

The correct option is A 133.3 g

On applying POAC for X and Y respectively, we get
$1\times n_X=1\times n_{X{Y}_3}$ ...(3)
$2\times n_{Y_2}=3\times n_{X{Y}_3}$ ...(4)
by (3) & (4), $n_{Y_2}=1.5\times n_X$
from (3) and (4)
$\frac{n_X}{2}=\frac{n_{Y_2}}{3}=\frac{n_{X{Y}_3}}{2}$
$\therefore$ Balanced equation : $2X+3Y_2\to 2X{Y}_3$
$n_{Y_2}=\frac{120}{60}\therefore n_{X{Y}_3}=\frac{4}{3}=1.333$
$\therefore m_{X{Y}_3}=\frac{4}{3}\times 100=133.3g$