Chemistry - Stoichiometry Question with Solution | TestHub

ChemistryStoichiometryBasic Methods of Calculations (POAC, LR)Medium2 minQB

ChemistryMediumsingle choice

Passage / Comprehension

We know that balancing a chemical equation is entirely based on the law of conservation of mass. However, the concept of the Principle of Atom Conservation (POAC) can also be related to the 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:

$A B C_{3} \to A B + C_{2}$

On applying POAC for A, B, and C and relating the three equations, we get:

$\frac{n _{A B C_{3}}}{2} = \frac{n _{A B}}{2} = \frac{n _{C_{2}}}{3}$ (n_x : number of moles of X)

Thus, the coefficients of ABC₃, AB, and C₂ in the balanced chemical equation will be 2, 2, and 3 respectively, and the balanced chemical equation can be represented as: Now answer the following questions:

$2 A B C_{3} \to 2 A B + 3 C_{2}$

Now answer the following questions:

If the atomic masses of X and Y are 10 & 30 respectively, then the mass $XY_{3}$ formed when 120 g of $Y_{2}$ reacts completely with X is:

Reaction $X + Y_{2} ⟶ X Y_{3}$ ⟶

Options:

Answer:

Solution:

On applying POAC for X and Y respectively, we get

$1 \times n_{X} = 1 \times n_{X Y_{3}}$ ....(i)

$2 \times n_{Y_{2}} = 3 \times n_{X Y_{3}}$ ......(ii)

by (i) and (ii)

$\frac{n _{X}}{2} = \frac{n _{Y_{2}}}{3} = \frac{n _{XY_{3}}}{2}$

∴ Balanced equation :

$2 X + 3 Y_{2} \to 2 XY_{3}$

$n_{Y_{2}} = \frac{120}{60} = 2 mol ∴ n_{X_{3}} = \frac{4}{3}$

$∴ m_{XY_{3}} = \frac{4}{3} \times 100 = 133.3 g$

Stream:JEESubject:ChemistryTopic:StoichiometrySubtopic:Basic Methods of Calculations (POAC, LR)

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...