Question

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remember: work on report questions alonet Data Analysis: A Based on the Mg and HCI used, calculate the theoretical yield of hydrogen gas for each trial Note that this is a limiting reactant calculation! Calculate the theoretical yield of hydrogen gas (in grams) for trial 1 1. Calculate the theoretical yield of hydrogen gas in (in grams) for trial 2 2. B. Calculate the grams of hydrogen actually produced in each trial (ie, the hydrogen measured in the tube). 1. Calculations for trial 1 2. Calculations for trial 2: C. For each trial calculate the percent yield of hydrogen gas Trial 2 Trial 1 0.0니00 9 0.0384 Mass of magnesium ribbon Volume of gas collected Temperature of the gas 15236 L4.35 733.S Barometric pressure R35 trr Vapor pressure of water 733 $1 to Partial pressure of hydrogen gas

Answer 1

Trial 1 :

Mass of Mg = 0.04 g

Volume of gas, V = 42.2 ml = 0.0422 lt

Temperature, T = 21.6 C = 21.6+273 = 294.6 K

Vapor pressure of water at 21.6 C = 19.35 torr

Barometric pressure of hydrogen gas = 752.86 torr

Pressure of dry hydrogen gas, P = 752.86 torr - 19.35 torr = 733.51 torr = 733.51 torr/760 = 0.965 atm

According to Ideal gas equation, PV = nRT

Ideal gas constant, R = 0.0821 lt atm/mol K

0.965 atm x 0.0422 lt = n x 0.0821 lt atm/mol K x 294.6 K

0.04073 mol /24.187 = n

Practical Number of moles, n = 0.001684 moles

Molar mass = 2.02 g/mol

Practical mass of hydrogen gas = No. of moles x molar mass = 0.001684 moles x 2.02 g/mol = 0.0034 gms

   Mg(s) + 2HCl(aq) -----> MgCl_2(S) + H_2(g)

Mass of Mg = 0.04 g

Molar mass of Mg = 24.305 g/mol

No. of moles of Mg = mass/molar mass = 0.04 g/24.305 g/mol = 0.00165 mol

Molar ratio of Mg and hydrogen gas = 1:1

Theoretical moles of hydrogen gas = 0.00165 mol

As molarity and volume of HCl is not mentioned. I considered Magnesium acts as limiting reactant

Molar mass of hydrogen gas = 2.02 g/mol

Theoretical weight of hydrogen gas = No. of moles x molar mass = 0.00165 mol x 2.02 g/mol = 0.00333 gms

Percent yield = (practical yield/theoretical yield) x 100 = 0.0034/0.00333 ) x 100 = 102.15 %

Trial 2 :

Mass of Mg = 0.0384 g

Volume of gas, V = 41.1 ml = 0.0411 lt

Temperature, T = 21.6 C = 21.6+273 = 294.6 K

Vapor pressure of water at 21.6 C = 19.35 torr

Partial pressure of hydrogen gas = 752.86 torr

Pressure of dry hydrogen gas, P = 752.86 torr - 19.35 torr = 733.15 torr = 733.15 torr/760 = 0.965 atm

According to Ideal gas equation, PV = nRT

Ideal gas constant, R = 0.0821 lt atm/mol K

0.965 atm x 0.0411 lt = n x 0.0821 lt atm/mol K x 294.6 K

0.039662 mol /24.187 = n

Practical Number of moles, n = 0.00164 moles

Molar mass = 2.02 g/mol

Practical mass of hydrogen gas = No. of moles x molar mass = 0.00164 moles x 2.02 g/mol = 0.0033 gms

   Mg(s) + 2HCl(aq) -----> MgCl_2(aq) + H_2(g)

Mass of Mg = 0.0384 g

Molar mass of Mg = 24.305 g/mol

No. of moles of Mg = mass/molar mass = 0.0384 g/24.305 g/mol = 0.00158 mol

Molar ratio of Mg and hydrogen gas = 1:1

Theoretical moles of hydrogen gas = 0.00158 mol

As molarity and volume of HCl is not mentioned. I considered Magnesium acts as limiting reactant

Molar mass of hydrogen gas = 2.02 g/mol

Theoretical weight of hydrogen gas = No. of moles x molar mass = 0.00158 mol x 2.02 g/mol = 0.00319 gms

Percent yield = (practical yield/theoretical yield) x 100 = 0.0033/0.00319 ) x 100 = 103.45 %