Question

The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ______.

Answer 1

1056

Answer 2

The overall reaction is:

$C_2H_5OH \longrightarrow CH_3CHO \longrightarrow CH_3COOH \longrightarrow CH_3COOC_2H_5 \longrightarrow CH_3CONH_2$

Moles of $CH_3CHO$ formed $= 16 \times 0.6 = 9.6$

Moles of $CH_3COOH$ formed $= 9.6 \times 0.8 = 7.68$

Moles of $CH_3COOC_2H_5$ formed $= 7.68 \times 0.5 = 3.84$

Moles of $CH_3CONH_2$ formed $= 3.84 \times 0.9 = 3.456$

Mass of $CH_3CONH_2 = 3.456 \times 59 = 203.904 g$

The second reaction is:

$C_2H_5OH \longrightarrow C_2H_5Br \longrightarrow C_2H_5OC_2H_5$

Moles of $C_2H_5Br$ formed $= 16 \times 0.7 = 11.2$

Moles of $C_2H_5OC_2H_5$ formed $= 11.2 \times 0.75 = 8.4$

Mass of $C_2H_5OC_2H_5 = 8.4 \times 74 = 621.6 g$

The third reaction is:

$C_2H_5OH \longrightarrow CH_2=CH_2 \longrightarrow CH_2Br-CH_2Br$

Moles of $CH_2=CH_2$ formed $= 16 \times 0.85 = 13.6$

Moles of $CH_2Br-CH_2Br$ formed $= 13.6 \times 0.4 = 5.44$

Mass of $CH_2Br-CH_2Br = 5.44 \times 188 = 1022.72 g$

Total mass of products $= 203.904 + 621.6 + 1022.72 = 1848.224 g$

The fourth reaction is:

$C_2H_5OH \longrightarrow CH_3CHO \longrightarrow CH_3COOH \longrightarrow (CH_3CO)_2O \longrightarrow CH_3COCl \longrightarrow CH_3CO-N(CH_3)_2$

Moles of $CH_3CHO$ formed $= 16 \times 0.6 = 9.6$

Moles of $CH_3COOH$ formed $= 9.6 \times 0.8 = 7.68$

Moles of $(CH_3CO)_2O$ formed $= 7.68 \times 0.5 = 3.84$

Moles of $CH_3COCl$ formed $= 3.84 \times 0.9 = 3.456$

Moles of $CH_3CO-N(CH_3)_2$ formed $= 3.456 \times 0.6 = 2.0736$

Mass of $CH_3CO-N(CH_3)_2 = 2.0736 \times 87 = 180.39 g$

Overall reaction:

$C_2H_5OH \longrightarrow CH_2=CH_2 \longrightarrow C_2H_4Br_2 \longrightarrow CH \equiv CH \longrightarrow AgC \equiv CAg \longrightarrow C + Ag$

$16 \longrightarrow 16 \times 0.85 \longrightarrow 16 \times 0.85 \times 0.4 \longrightarrow 16 \times 0.85 \times 0.4 \times 0.6 \longrightarrow 16 \times 0.85 \times 0.4 \times 0.6 \times 0.7 \longrightarrow 16 \times 0.85 \times 0.4 \times 0.6 \times 0.7 \times 0.8$

Moles of $Ag = 16 \times 0.85 \times 0.4 \times 0.6 \times 0.7 \times 0.8 = 1.8304$

Mass of $Ag = 1.8304 \times 108 = 197.6832 g$

Moles of $C = 1.8304 / 2 = 0.9152$

Mass of $C = 0.9152 \times 12 = 10.9824 g$

Moles of $S = 16 \times 0.75 \times 0.8 \times 1.1 = 10.56$

Mass of $S = 10.56 \times 100 = 1056$