Question

Please help with these 2 worksheets!

I especially am struggling with part D.

Trial 1 Trial 2 Mass of magnesium ribbon 0.04219 Volume of gas collected 0.03869 37 ml 21.0°C Temperature of the gas 43ml 21.1C 29.23inHg Barometric pressure 29.23in its 18.77 torr → 0.71intig 28.509 in Hg Vapor pressure of water 18.65 torr +0.717 into Partial pressure of hydrogen gas 28.513 in Hg Lab Partner: Aya Siblani IV. QUESTIONS and CALCULATIONS A. Calculate the moles of hydrogen gas actually produced in each trial. How many grams is this? 1. Calculations for trial 1. 2. Calculations for trial 2.

Answer 1

Magnesium (Mg) reacts with steam to liberate H2 (g) and form MgO. Note that it will not react with cold water.

Mg(s)+H,00) 100°C > MgO (6) +H2(8)

Part A

Number of moles of Hydrogen gas can be calculated using Ideal gas equstion: PV = nRT

So, number of moles, n= PV/RT

Trial 1

Pressure (P) = 28.513 in Hg = 28.513×0.0330 atm = 0.94 atm

Volume of gas (V) = 37 ml = 0.037 L

Temperature (T)= (21+273) K = 294 K

Ideal gas constant, R = 0.082 atm.L/mol.K

Number of moles, n= (0.94 atm×0.037 L)/( 0.082 atm.L/mol.K × 294 K )

ð n= 1.44×10-3 moles

ð mass= Number of moles× molar mass = 1.44×10-3 moles× 2 gm/mole = 2.88 mg

Trial 2

Pressure (P) = 28.509 in Hg = 28.509×0.033 atm = 0.94 atm

Volume of gas (V) = 43 ml = 0.043 L

Temperature (T)= (21.1+273) K = 294.1 K

Ideal gas constant, R = 0.082 atm.L/mol.K

Number of moles, n= (0.94 atm×0.043 L)/( 0.082 atm.L/mol.K × 294.1 K )

ð n= 1.67×10-3 moles

ð mass= Number of moles× molar mass = 1.67×10-3 moles× 2 gm/mole = 3.35 mg

Part D

Q1: Mole fraction of water vapor in sample can be calculated from Dalton’s law of Partial pressure. According to Dalton's law of partial pressures, the total pressure by a mixture of non-reacting gases is equal to the sum of the partial pressures of each of the constituent gases.

The mole fraction of a specific gas in a mixture of gases is equal to the ratio of the partial pressure of that gas to the total pressure exerted by the gaseous mixture.

So, mole fraction of water vapour= vapour pressure of water/barometric pressure = 0.717/29.23 = 0.0245

Q2: Let mole fraction of water vapour be y

mole fraction of water vapour = moles of water vapour/( moles of water vapour+ moles of hydrogen gas)

So, 0.0245= y/(y+1.67×10-3)

Solving this we get, moles of water vapour= 4×10-5 moles

Q3: Number of molecules of water = moles of water vapour×Avogadro number= 4×10-5 × 6×1023 = 2.455×1019 molecules

Q4: mass of gas mixture = mass of Hydrogen gas + mass of water vapor= 2.88 mg + (4×10-5 moles×18000 mg/mol) = 3.6 mg

Density= mass of gas mixture/ volume = 3.6mg/37 ml= 0.097 gm/L