When bar-headed geese fly at very high altitudes (possibly over Mount Everest!), they breathe very thin air where the partial pr
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Answer:
0.9483 grams of manganese dioxide should be added to excess HCl.
Explanation:
Pressure of the chlorine gas = P = 795 Torr = 1.046 atm (1 atm = 760 Torr)
Volume of the chlorine gas = V = 255 ml = 0.255 L
Temperature of the chlorine gas = T = 25°C= 298.15 K
Moles of chlorine gas = n
Using ideal gas equation:
PV = nRT
n = 0.01090 mol
According to reaction , 1 mole of chlorine gas is obtained from 1 mole of manganese dioxide.
Then 0.01090 moles of chlorine gas will be obtained from:
manganese dioxide
Mass of 0.01090 moles of manganese dioxide:
0.01090 mol × 87 g/mol = 0.9483 g