CHM 147 Oakland University Department of Experiment 6 How Much Is Too Much? Data

Question

CHM 147 Oakland University Department of Experiment 6 How Much Is Too Much? Data Sheet (2 of 2 Name: Date: Your Partner: Names of other students you worked with Am peaa ce Part C: Determination of the percent zinc in a zinc/potassium chloride mixture Unknown number M 2 a Trial #1 Trial #2 (your data) (other group's data) Mass of Total Sample I. S 32 (grams) Total Volume of H 2(g cd o mi 2 Sam L volume of HCI used (mL) lo o M Atmospheric pressure ng .3 in (Make sure to record units) Atmospheric pressure in mmH mil water temperature 1 c Vapor Pressure of water at measured temperature s i mmv HS (This data can be found in CRC handbook found in the lab or online. Be sure to record units. This value may need to be converted to mmHg) Vapor pressure of water in mmHg ST12 H Rev. 8-25-15

Explanation / Answer

Data for the Triel-1

1.533 g of Zn and KCl mixture

Total volume of H2 (gas) = 340mL

Volume of HCl used= 10mL

Atmospheric pressure= 719mmHg

Water temperature = 12.0 degree Celsius = 273.15+12.0 = 285.15K

Vapor pressure of water at temperature = 10.518 mm Hg

Pressure of H2 (gas) = 719mmhg - 10.518 mm hg = 708.482 mmHg

R = 62.364 L-mmHg/ºK-mol (Gas constant in mmHg K1 mol1)

The zinc will react with the HCl to produce ZnCl2 and H2 gas:

2HCl + Zn ----> ZnCl2 + H2

The moles of Zn equal the moles of H2 gas.

Find the moles of H2 gas

n = PV/RT.

n = 708.482*0.340/(62.364* 285.15)

     = 0.0135456 mol

The moles of zinc = 0.0135456 mol

Molar mass of Zn = 65.38 g/mol

Mz = 0.0135456mol x 65.38 g/mol = 0.886 g

Now find the percent in the mixture from Mz/1.533 * 100

Zn% = (0.886 g /1.533g) x 100

Zn % = 57.79

Data for the Triel-2

1.650 g of Zn and KCl mixture

Total volume of H2 (gas) = 250mL

Volume of HCl used= 10mL

Find the moles of H2 gas

n = PV/RT.

n = 708.482*0.250/(62.364* 285.15)

     = 0.00996 mol

The moles of zinc = 0.00996 mol

Molar mass of Zn = 65.38 g/mol

Mz = 0.00996 mol x 65.38 g/mol = 0.651g

Now find the percent in the mixture from Mz/1.650 * 100

Zn% = (0.651 g /1.650g) x 100

Zn % = 39.45

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