Q. 7.57

Question

Carbon disulfide and carbon monoxide are produced when carbon is heated with sulfur dioxide.

a. How many moles of $C$ are needed to react with $0.500 m o l e$ of $S O_{2}$ ?

b. How many moles of $C O$ are produced when $1.2$ moles of $C$ reacts?

c. How many moles of $S O_{2}$ are needed to produce $0.50$ mole of $C S_{2}$ ?

d. How many moles of $C S_{2}$ are produced when $2.5$ moles of $C$ reacts?

Step-by-Step Solution

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Answer

a. The number of moles of $C$ needed is $1.25$ mole.

b. The number of moles of $C O$ produced is $0.96$ mole.

c. The number of moles of $S O_{2}$ needed is $1$ mole.

d. The number of moles of $C S_{2}$ produced is $0.5$ mole.

1Part(a) Step 1: Given Information

We need to find the moles of $C$ which is needed to react with $0.500 m o l e$ of $S O_{2}$ .

2Part (a) Step 2: Calculation

Using mole factor

$m o l e f c t o r = \frac{5 m o l e C}{2 m o l e S O_{2}}$

$5 m o l e C = 2 m o l e S O_{2}$

according to question

$(0.5 m o l e S O_{2}) \times (\frac{5 m o l e C}{2 m o l e S O_{2}}) = 1.25 m o l e C$

3Part (b) Step 1: Given Information

We need to find the number of moles of $C O$ that are produced when $1.2$ moles of $C$ reacts.

4Part (b) Step 2: Calculation

Using mole factor

$m o l e f a c t o r = \frac{4 m o l e C O}{5 m o l e C}$

according to question

$5 m o l e C = 4 m o l e C O (1.2 m o l e C) \times (\frac{4 m o l e C O}{5 m o l e C}) = 0.96 m o l e C O$

5Part (c) Step 1: Given Information

We need the find the number of moles of $S O_{2}$ that needed to produce $0.50$ mole of $C S_{2}$ .

6Part (c) Step 2: Calculation

Using the mole factor

$m o l e f a c t o r = \frac{2 m o l e S O_{2}}{1 m o l e C S_{2}}$

according to question

$2 m o l e S O_{2} = 1 m o l e C S_{2} (0.5 m o l e C S_{2}) \times (\frac{2 m o l e S O_{2}}{1 m o l e C S_{2}}) 1 m o l e S O_{2}$

7Part (d) Step 1: Given Information

We need to find the number of moles $C S_{2}$ that are produced when $2.5$ moles of $C$ reacts.

8Part (d) Step 2: Calculation

Using the mole factor

$m o l e f a c t o r = \frac{1 m o l e C S_{2}}{5 m o l e C}$

according to question

$5 m o l e C = 1 m o l e C S_{2} (2.5 m o l e C) \times (\frac{1 m o l e C S_{2}}{5 m o l e C}) = 0.5 m o l e C S_{2}$

Q. 7.55

Q. 7.60

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