Question

Carbon disulphide is produced as per following reaction : 5C + 2 502 → CS2 + 4CO 90 kg of coke reacts with required amount of SO2. If percentage yield of the reaction is 40% then the mass of CS2 formed is​

Answer 1

Answer:

219

Explanation:

The molar mass of SO2 is 32+2(16)=64 g/mol

The molar mass of CS2 is 12+2(32)=76 g/mol

2 moles of SO2 form 1 mole of CS2.

2×64=128 g of SO2 form 76 g of CS2.

450 kg of SO2 will form 12876×450=267 kg of CS2.

However, the conversion is only 82%.

Hence, 10082×267=219 kg of CS2 will be formed.

Answer 2

Given:

$5 \mathrm{C}+2 \mathrm{SO}_{2} \rightarrow \mathrm{CS}_{2}+4 \mathrm{CO}$

90 kg of coke reacts with required amount of SO2

percentage yield of the reaction is 40%

• $\text { The molar mass of } \mathrm{SO}_{2} \text { is } 32+2(16)=64 \mathrm{~g} / \mathrm{mol}$

• $\text { The molar mass of } \mathrm{CS}_{2} \text { is } 12+2(32)=76 \mathrm{~g} / \mathrm{mol}$

$2 \text { moles of } \mathrm{SO}_{2} \text { form } 1 \text { mole of } \mathrm{CS}_{2}$

$2 \times 64=128 \mathrm{~g} \text { of } \mathrm{SO}_{2} \text { form } 76 \mathrm{~g} \text { of } \mathrm{CS}_{2}$

$90 \mathrm{~kg} \text { of } \mathrm{SO}_{2} \text { will form } \frac{76}{128} \times 90$

$=0.59 \times 90\\\\=53\mathrm{kg} \text { of } \mathrm{CS}_{2}$

$\text { However, the conversion is only } 40 \% \text {. }$

$\frac{40}{100} \times 53\\\\=0.4\times53\\\\=21\mathrm{~kg} \text { of } \mathrm{CS}_{2} \text { will be formed. }$

Hence 21 kg of  $\mathrm{CS}_{2}$  will be formed.