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Answer:
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of chlorine gas =
1 atm = 760 Torr
V = Volume of chlorine gas = 385 mL = 0.385 L ( 1 mL - 0.001 L)
n = number of moles of chlorine gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of chlorine gas =25 °C= 25 + 273 K = 300 K
Putting values in above equation, we get:
According to reaction, 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide ,then 0.01573 moles of chlorine gas will be obtained from :
of manganese dioxide
Mass of 0.01573 moles of manganese dioxide:
0.01573 mol × 86.94 g/mol = 1.368 g
1.368 grams of manganese dioxide must b added to HCl to obtain 385 mL of chlorine gas.