Question

Q 18/45 Which of the following contains maximum
number of moles?
16 g of SO3
22 g of N20
O
32 g of 03
O
10 g of Cl2​

Answer 1

your ans is in attachment

THE ANS IS 32 g OF 03

Answer 2

Answer:

Out of the given compounds, the maximum number of moles contains $32g$ of $O_{3}$.

Therefore, option c) $32g$ of $O_{3}$ is correct.

Step-by-step explanation:

The number of moles = $\frac{Given mass}{Molar mass}$

Case 1) $16g$ of $SO_{3}$.

• The given mass = $16g$
• The molar mass of $SO_{3}$ = sulfur's atomic mass + 3 × oxygen's atomic mass.
• The molar mass of $SO_{3}$ = $32 + 3\times 16$ = $80g/mol$

Therefore, the number of moles = $\frac{16}{80}$ = $0.2mol$.

Case 2) $22g$ of $N_{2}O$.

• The given mass = $22g$
• The molar mass of $N_{2}O$ = 2 × nitrogen's atomic mass + oxygen's atomic mass.
• The molar mass of $N_{2}O$ = $2\times 14 +16$ = $44g/mol$

Therefore, the number of moles = $\frac{22}{44}$ = $0.5mol$.

Case 3) $32g$ of $O_{3}$.

• The given mass = $32g$
• The molar mass of $O_{3}$ =  3 × oxygen's atomic mass.
• The molar mass of $O_{3}$ = $3\times 16$ = $48g/mol$

Therefore, the number of moles = $\frac{32}{48}$ = $0.6mol$.

Case 4) $10g$ of $Cl_{2}$.

• The given mass = $10g$
• The molar mass of $Cl_{2}$ = 2 × chlorine's atomic mass.
• The molar mass of $Cl_{2}$ = $2\times 35$ = $70g/mol$

Therefore, the number of moles = $\frac{10}{70}$ = $0.1 mol$.

Hence, the maximum number of moles contains $32g$ of $O_{3}$.