7. How many liters of chlorine gas (70.91 g/mol) at 25°C and 0.950 atm can be pr

Question

7. How many liters of chlorine gas (70.91 g/mol) at 25°C and 0.950 atm can be produced by the reaction of 12.0 g of MnO2 (86.94 g/mol) with excess HCI(aq) according to the following chemical equation? MnO2(s) + 4HCI(aq) MnCl2(aq) + 2H2O(l) + Cl2(g) 8. Hydrogen gas was produced in the laboratory and was collected over water at 25°C. The total pressure of the hydrogen gas and water vapor mixture was 742.5 mm Hg. What is the partial pressure of the hydroge gas if the vapor pressure of water at 25 C is 23.8 mm Hg? BONUS d and the temperature was tripled what is the final pressure?

Explanation / Answer

7) moles of MnO2 = 12 g / 86.94 == 0.138 moles

1 mole Mno2 gives 1 mole of Cl2

so, moles of Cl2 = 0.138 moles from reaction.

from, PV=nRT

V = nRT /P

= 0.138 moles x 0.082 x 298.15 / 0.950 atm

= 3.552 L

8)

Total pressure = 742.5 mm Hg

pressure of water vapor = 23.8 mm Hg

Pressure of hydrogen = 742.5 - 23.8 =718.7 mm Hg

bonus

Start with

P1V1 / T1 = P2V2 / T2

P2 = P1 x(V1/V2) x (T2/T1)

= 1 atm x ( 1/2) x (3/1)

= 1.5 atm

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