1st calculate the number of mol of Cl2
Given:
P = 775.0 torr
= (775.0/760) atm
= 1.0197 atm
V = 125.0 mL
= (125.0/1000) L
= 0.125 L
T = 25.0 oC
= (25.0+273) K
= 298 K
find number of moles using:
P * V = n*R*T
1.0197 atm * 0.125 L = n * 0.08206 atm.L/mol.K * 298 K
n = 5.213*10^-3 mol
From reaction,
Mol of MnO2 reacted = mol of Cl2 formed
= 5.213*10^-3 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
= 5.213*10^-3 mol * 86.94 g/mol
= 0.4532 g
Answer: 0.453 g
Question 20 of 34 > Chlorine can be prepared in the laboratory by the reaction of...
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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 MnO,(s) + 4 HCl(aq) — MnCl, (aq) + 2 H,O(1) + Cl, (g) How much MnO,(s) should be added to excess HCl(aq) to obtain 315 mL CI, (g) at 25°C and 775 Torr? mass of MnO,
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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 MnO,(s) + 4 HCl(aq) + MnCl(aq) + 2H,0(1) + Cl (8) How much. Mno,(s) should be added to excess HCl(aq) to obtain 155 mL CI(8) at 25 °C and 1.00 x 10 kPa? mass: g MnO Question Source MRG-General Chemistry | Publisher
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