calculate the mass of MnO2 needed to produce 1.78 litres of chlorine gas at STP according to the equation
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
[H = 1, Cl = 35.5, Mn = 54.9]
The equation is
MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
1 mole of MnO2 = 1 mole of Cl2
86.9g of MnO2 = 22.4 L of Cl2 at NTP
∴ 1.78 L Cl2 = 1.78×86.922.4 = 6.905 g of MnO2