Respuesta :
Answer:
13.3 grams
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
Answer:
We need 13.9 grams of CS2
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
