Respuesta :
Answer:
929 mg of MnO₂ should be added to excess HCl, in order to obtain 385 mL of Cl₂
Explanation:
The reaction states:
MnO₂ (s) + 4 HCl (aq) → MnCl₂(aq) + 2H₂O (l) + Cl₂(g)
First of all we need the moles of produced chlorine. We can find it out by the Ideal Gases Law → P . V = n . R . T
- We need to convert the 795 Torr to atm → 795 Torr . 1 atm / 760 Torr = 1.05 atm
- We need to convert the 385 mL to L → 385 mL . 1L / 1000mL = 0.385L
- We need T° K = 25°C + 273 = 298K
1.05 atm . 0.385L = n . 0.082 . 298K
n = 1.05 atm . 0.385L / 0.082 . 298K → 0.0165 moles of Cl₂
Ratio is 1:1. 1 mol of chlorine gas can be made by 1 mol of MnO₂
Therefore, If I produced 0.0165 moles of Cl₂ I definitely used 0.0165 moles of MnO₂
We convert the moles to mass → 0.0165 mol . 86.94 / 1 mol = 1.43 g
We can convert the g to mg → 1.43 g . 1000 mg / 1g = 1430 mg
Answer:
We need 1.43 grams of MnO2
Explanation:
Step 1: Data given
Volume of Cl2 = 385 mL
Temperature = 25 °C = 298 K
Pressure = 795 torr = 1.04605 atm
molar mass of MnO2 = 86.93 g /mol
Step 2: The balanced equation
MnO2(s) + 4HCl(aq) ⟶ MnCl2 (aq) + 2H2O (l) + Cl2 (g)
Step 3: Calculate moles Cl2
p*V = n*R*T
n = (p*V)/(R*T)
⇒n = number of moles Cl2 = TO BE DETERMINED
⇒p = the pressure of Cl2 = 1.04605 atm
⇒V = the volume of Cl2 = 0.385 L
⇒R = the gas constant = 0.08206 L*atm/mol*K
⇒T = the temperature = 25 °C = 298 K
n = (1.04605 * 0.385)/(0.08206 * 298)
n = 0.0165 moles
Step 4: Calculate moles MnO2
For 1 mol MnCl2, 2 moles H2O and 1 mol Cl2 produced, we need 1 mol MnO2 and 4 moles HCl
For 0.0165 moles Cl2 we need 0.0165 moles MnO2
Step 5: Calculate mass of MnO2
Mass MnO2 = moles MnO2 * molar mass MnO2
Mass MnO2 = 0.0165 moles * 86.93 g/mol
Mass MnO2 = 1.43 grams
We need 1.43 grams of MnO2
