## 8.45 Preparation of Carbon Disulfide

Carrie Huang 2F
Posts: 13
Joined: Fri Sep 25, 2015 3:00 am

### 8.45 Preparation of Carbon Disulfide

Problem 8.45 reads: Carbon disulfide can be prepared from coke and elemental sulfur:

4C(s)+S8(s) --> 4CS2(l) deltaH=+358.8kJ

(a) How much heat is absorbed in the reaction of 1.25 mol S8 at constant pressure?
(b) Calculate the heat absorbed in the reaction of 197g of carbon with an excess of sulfur.
(c) If the heat absorbed in the reaction was 415kJ, how much CS2 was produced?

Can someone walk me through the entire process of solving this, along with explanations? Thanks.

Serena Zhang 3D
Posts: 28
Joined: Fri Sep 25, 2015 3:00 am

### Re: 8.45 Preparation of Carbon Disulfide

a). You know from the balanced equation that +358.8kJ is absorbed for a reaction involving 1 mol of $S_{8}$. To find how much heat is absorbed for a reaction where 1.25 mol of $S_{8}$ goes to completion, multiply 358.8kJ by 1.25.
b). To calculate the heat absorbed we need to know how many moles of C there are. Since the problem mentions there is an excess of sulfur, C is the limiting reagent. Divide 197g of C by the molar mass to obtain the moles of C. From the balanced equation you can see that for every 4 moles of C consumed in the reaction, 358.8kJ is absorbed. Multiply the moles of C by 358.8kJ/4mol to get the answer.
c). Calculate how many moles of $CS_{2}$ were produced by solving 415kJ/358.8kJ, then multiplying that quantity by 4 because remember 4 moles of $CS_{2}$ are produced in a reaction that absorbs 358.8kJ. Multiply the moles of $CS_{2}$ by the molar mass to get grams of $CS_{2}$ produced.

Leonardo Le Merle 1D
Posts: 56
Joined: Wed Feb 27, 2019 12:16 am

### Re: 8.45 Preparation of Carbon Disulfide

a) (1.25mol S8) * (358.8 kJ/1mol S8) = 448.5 kJ
b) (197g C) * (1mol C/12.01g C) * (358kJ/4 mol C) = 1471.35 kJ
c) (415 kJ/358.8kJ) = (x mol CS2/4mol CS2); x = 4.63mol CS2

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