Question

счасоп.
Example 5.21 Calculate the weight of MnO2 required to
produce 1.50 L of chlorine at 27°C and 1.50 atm pressure according to
the following equation :
Mno2 + 4HCI - MnCl2 + 2H2O + CI2​

Answer 1

Answer:

95.7 g

Explanation:

Mno2 + 4HCI → MnCl2 + 2H2O + CI2

observe the co-efficients....

1 mole MnO2 → 1 mole Cl2

(x/55+2×16) → 22.4 L ( mole = W/M)

x = 1948.8 g

means....at STP...

22.4 L of Cl2 → 1949g MnO2

now....at... T' = 27°C ; P = 1.5atm ; V = 1.5L ; V' = ?

since PV/T = P'V'/T'

V' = PVT'/P'T'

hence V' = 1 × 1.5 × 300/1.5×273 = 1.1L

now....

1.1 L of Cl2 → ? g MnO2

MnO2 = 1.1 × 1949/22.4 = 95.7g