chlorine can be prepared by in the laboratory by the reaction of manganese dioxide with the...
Homework Answers
Given:
P = 795.0 torr
= (795.0/760) atm
= 1.0461 atm
V = 345.0 mL
= (345.0/1000) L
= 0.345 L
T = 25.0 oC
= (25.0+273) K
= 298 K
find number of moles using:
P * V = n*R*T
1.0461 atm * 0.345 L = n * 0.08206 atm.L/mol.K * 298 K
n = 1.476*10^-2 mol
From reaction,
Mol of MnO2 reacted = mol of Cl2 produced
= 1.476*10^-2 mol
Molar mass of MnO2,
MM = 1*MM(Mn) + 2*MM(O)
= 1*54.94 + 2*16.0
= 86.94 g/mol
use:
mass of MnO2,
m = number of mol * molar mass
= 1.476*10^-2 mol * 86.94 g/mol
= 1.283 g
Answer: 1.28 g
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chlorine can be prepared by in the laboratory by the reaction of
manganese dioxide with the...
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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s) should be added to excess HCl(aq) to obtain 285 mL Cl2(g) at 25 °C and 805 Torr? mass of MnO2: g
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