Question

How many grams of carbon disulfide are needed to completely consume 17.9 L of chlorine gas according to the following reaction at 25 °C and 1 atm? carbon disulfide ( s ) + chlorine ( g ) carbon tetrachloride ( l ) + sulfur dichloride ( s )

Answer 1

We need 13.9 grams of CS2

Step-by-step explanation:

Step 1: Data given

Volume of chlorine gas produced = 17.9 L

Temperature = 25°C = 298 K

Pressure = 1atm

carbon disulfide = CS2

chlorine = Cl2

carbon tetrachloride = CCl4

sulfur dichloride = SCl2

Step 2: The balanced equation

CS2(s) + 4Cl2(g) → CCl4(l) + 2SCl2(s)

Step 3: Calculate moles Cl2

p*V = n*R*T

⇒with p = the pressure of chlorine gas = 1 atm

⇒with V = the volume of chlorine gas = 17.9 L

⇒with n = the number of moles chlorine gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = the temperature = 298 K

n =(p*V) / (R*T)

n = (1 * 17.9) / (0.08206 * 298)

n= 0.732 moles

Step 4: Calculate moles CS2

For 1 mol CS2 we need 4 moles Cl2 to produce 1 mol CCl4 and 2 moles SCl2

For 0.732 moles Cl2 we'll have 0.732/4 = 0.183 moles

Step 5: Calculate mass CS2

Mass CS2 = moles CS2 * molar mass CS2

Mass CS2 = 0.183 moles * 76.14 g/mol

Mass CS2 = 13.9 grams

We need 13.9 grams of CS2

Answer 2

13.3 grams

Step-by-step explanation:

Given the reaction:

CS2 (s) + 4 Cl2 (g) -> CCl4 (l) + 2 SCl2 (s)

Data

- Volume, V = 17.9 L

- Temperature, T = 25 °C or 25 + 273 = 298 K

- Pressure, P = 1 atm

- gas constant, R = 0.082 atm*L/(mol*K)

From the ideal gas law:

P*V = n*R*T

n = P*V/(R*T)

n = 1*17/(0.082*298)

n = 0.7 mol of Cl2

From the reaction we know that 1 mol of CS2 reacts with 4 moles of Cl2, then 0.7 mol of Cl2 reacts with x moles of CS2:

1 mol of CS2 / x moles of CS2 = 4 moles of Cl2 / 0.7 mol of Cl2

x = 1*0.7/4

x = 0.175 mol of CS2

The molecular weight of CS2 is 12 + 2*32 = 76 grams/mol

Then, 0.175 mol are equivalent to 0.175*76 = 13.3 grams