5 g of benzene on nitration gave 6.6 g of nitrobenzene. The theoretical yield of the nitrobenzene will be

Question

5 g of benzene on nitration gave 6.6 g of nitrobenzene. The theoretical yield of the nitrobenzene will be (A) 4.5 g (B) 5.6 g (C) 8.09 g (D) 6.6 g

Tardigrade Admin · Accepted Answer

underset78 g C 6 H 6+ HNO 3 longrightarrow underset123 g C 6 H 5 NO 2+ H 2 O Now since 78 g of benzene on nitration give =123 g nitrobenzene hence 5 g of benzene on nitration give =(123/78) × 7.88 g nearest answer is (c) i.e. theoritical yield =7.88 g

Q. 5 g of benzene on nitration gave 6.6 g of nitrobenzene. The theoretical yield of the nitrobenzene will be

4.5 g

5.6 g

8.09 g

6.6 g

$78 g C_{6} H_{6} + H N O_{3} ⟶ 123 g C_{6} H_{5} N O_{2} + H_{2} O$
Now since $78 g$ of benzene on nitration give $= 123 g$ nitrobenzene hence $5 g$ of benzene on nitration give $= \frac{123}{78} \times 7.88 g$
nearest answer is (c) i.e. theoritical yield $= 7.88 g$

Q. 5 g of benzene on nitration gave 6.6 g of nitrobenzene. The theoretical yield of the nitrobenzene will be

4.5 g

5.6 g

8.09 g

6.6 g

78gC6​H6​​+HNO3​⟶123gC6​H5​NO2​​+H2​O Now since 78g of benzene on nitration give =123g nitrobenzene hence 5g of benzene on nitration give =78123​×7.88g nearest answer is (c) i.e. theoritical yield =7.88g

$78 g C_{6} H_{6} + H N O_{3} ⟶ 123 g C_{6} H_{5} N O_{2} + H_{2} O$
Now since $78 g$ of benzene on nitration give $= 123 g$ nitrobenzene hence $5 g$ of benzene on nitration give $= \frac{123}{78} \times 7.88 g$
nearest answer is (c) i.e. theoritical yield $= 7.88 g$