C_2H_4 can react with H_2 in presence of a catalyst to form C_2H_6 as per the following reaction, C_2H_4g+H_2 g⟶ Catalyst C_2H_6g The amount of C_2H_4 in grams required to produce 50 grams of C_2H_6 is

Correct answer: (C) 46.67 g. Given reaction C_2H_4g+H_2 g⟶ Catalyst C_2H_6g C_2H_4 molecular weight = 28 , C_2H_6 molecular weight =30 , ⇒For the form 30 gm C2H6 required C2H4 =28 ⇒For50 the form 50 gm C2H6 required C2H4 = (28×50)/30=46.67g

Q10 / 160 · TS EAMCET 2020 (09 Sep Shift 1) · Chemistry

$C_{2} H_{4}$ can react with $H_{2}$ in presence of a catalyst to form $C_{2} H_{6}$ as per the following reaction, $C_{2} H_{4} (g) + H_{2} (g) \overset{Catalyst}{⟶} C_{2} H_{6} (g)$ The amount of $C_{2} H_{4}$ in grams required to produce $50$ grams of $C_{2} H_{6}$ is

A
$36.44 g$
B
$22.18 g$
Correct answer
C
$46.67 g$
D
$57.11 g$

Solution

Given reaction
$C_{2} H_{4} (g) + H_{2} (g) \overset{Catalyst}{⟶} C_{2} H_{6} (g)$

$C_{2} H_{4} m o l e c u l a r w e i g h t = 28, C_{2} H_{6} m o l e c u l a r w e i g h t = 30,$

$\Rightarrow$ For the form $30$ gm C₂H₆required C₂H₄= $28$

$\Rightarrow$ For $50$ the form 50 gm C₂H₆required C₂H₄= $\frac{28 \times 50}{30} = 46.67$ g

Hence, the correct answer is option: 'C'.

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C2H4 can react with H2 in presence of a catalyst to form C2H6 as per the following reaction, C2H4g+H2 g⟶ Catalyst C2H6g The amount of C2H4 in grams required to produce 50 grams of C2H6 is

Solution

Relevant Questions

$C_{2} H_{4}$ can react with $H_{2}$ in presence of a catalyst to form $C_{2} H_{6}$ as per the following reaction, $C_{2} H_{4} (g) + H_{2} (g) \overset{Catalyst}{⟶} C_{2} H_{6} (g)$ The amount of $C_{2} H_{4}$ in grams required to produce $50$ grams of $C_{2} H_{6}$ is