Please help me answer these! Here is some information: mass of zinc used (g) 0.5

Question

Please help me answer these!

Here is some information:

mass of zinc used (g) 0.500 g

volume of 6 M HCl used (mL) 10 mL

pressure in the flask before adding the gas syringe (atm) 1.00 atm

volume of hydrogen gas produced (mL) 185.6 mL

What was the expected volume of hydrogen produced in the reaction in part 1? Compare your result with the value you measured in the lab and calculate the percent error using the formula below.

What was the expected volume of hydrogen produced in the reaction in part 2? Compare your result with the value you measured in the lab and calculate the percent error using the formula below.

Compare the experimental value for the molar volume at 21.5 ºC obtained in part 2 with the value listed in the Background section of the lab manual. The molar mass of Zn is 65.38 g/mol. Calculate the experimental error according to the equation below.

Calculate and compare the value for the molar volume at 21.5 °C obtained in part 2 with the value for the molar volume at 21.5 °C obtained in part 1. The molar mass of Zn is 65.38 g/mol.

How did using twice the amount of Zn in the second part of the experiment impact the molar volume? Explain your answer. From the data recorded in the table below, calculate the pressure that would be observed. temperature 21.5 ºC moles H2 volume 125.0 mL

moles of zinc reacted .00765 moles of hydrogen produced .00765 molar volume of the hydrogen gas at room temperature 24.12

Explanation / Answer

mass of zinc = 0.5 g; mmol of HCL = MV = 10*6 = 60 mmol of HCl, P = 1 atm, Vproduced = 186.6 mL

A)

expected volumeo f H2 in PART 1.

Zn + 2HCl --> H2 + ZnCl2

mol of Zn = mass/MW = 0.5/65.38 = 0.0076475 mol of Zn = 0.0076475*10^3 mmol = 7.64 mmol

mmol of HCl = 60 mmol

there is excess HCl, so

1 mol of Zn = 1 mol of H2

7.64 mmol of Zn = 7.64 mmol of H2 expected

Volume, from ideal gas law

PV = nRT

V = nrT/P = (7.64 *10^-3)(0.082)(21.5+273)/(1) = 0.1844 L = 184.4 mL expected

B)

from part 2 --> 0.00765 mol of H2 produces, so

V = mol * Vmolar = 0.00765 *24.12 = 0.1845 L = 184.5 mL of H2

C)

Compare:

Error% = (184.4 - 184.5 ) / (184.4 ) * 100% = -0.05422 % error, meaning that the calcualtion is UNDER what we got

D)

using twice of Zinc amonut will increase the Volume, but also the moles

therefore, the molar volume, i.e. = Volume / mol

remains the SAME, that is, it must be constant, since this is an intensive property

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