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# Lab: Molar Volume of Hydrogen Gas

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Part 1
The following are the givens from the experiment:
Initial Experiment with 10mL- 6 M HCl and 0.25g Zinc
Initial Teperature = 21.5C
Initial Pressure = 1.00atm
After adding 0.25g Zinc - The solution reacted
Pressue = 1.69atm
Temperature = 27.3C
Volume in Syrenge = 92.2mL

Part 2
Repeated Experiment with 10mL- 6 M HCl and 0.5g of Zinc
Initial Temperature = 21.5C
Initial Pressure = 1.00atm
After adding 0.5g Zinc -The solution reacted
Pressure = 2.41atm
Temperature =33.4C
Volume in Syrenge = 184.5mL

The following answers are what I have so far: Need to know if A,B,C,D,E, & F are correct for 1 & 2 if not please correct.

1. Record and calculate the following for the first half of the experiment:
a. Mass of zinc used (g) = 0.25g
b. Volume of 6M HCl used (mL) = 10mL
c. Molecular weight of zinc (g/mol) = 65.409 g/mol
d. Moles of zinc reacted = 0.25 g / 65.41 g/mol= 0.00382 moles
e. Moles of hydrogen produced = 0.00382 moles
f. Molar volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed as L/mol at X degrees C and a pressure of 1 atmosphere = 22.4 L/mole * 0.00382 moles = 0.0856 L or 85.6mL

2. For the second part of the experiment, everything was the same except that twice as much Zn was used. Record and calculate the following for this second reaction:
a. Mass of zinc used (g) = 0.50
b. Volume of 6M HCl used (mL) = 10mL
c. Molecular weight of zinc (g/mol) = 65.409 g/mol
d. Moles of zinc reacted = 0.50 g / 65.41 g/mol=0.00764
e. Moles of hydrogen produced = 0.00764 moles
f. Molar volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed as L/mol at X degrees C and a pressure of 1 atmosphere = 22.4 L/mole * 0.00764 moles = 0.1711 L

Also need the following answered from the above information: Numbers 3, 4, & 5

3. Compare the molar volumes obtained in the two parts of the experiment. What difference did using twice the amount of Zn in the second part?

4. Which is the limiting reactant - zinc or HCl? I think HCl

5. Compare the experimental value for the molar volume at 21C with the value listed in the Background section of the lab manual. Calculate the experimental error according to: % error = | experimental molar volume - listed molar volume | / (listed molar volume) * 100%

Background

Robert Boyle found that, under isothermal conditions (constant temperature), the pressure of a gas is inversely proportional to its volume. This also means that the product of the pressure and volume is a constant. The equations that describe this relationship are
P × V = constant
and
P1 × V1 = P2 × V2
Jacques Charles found that the volume of a gas under isobaric conditions (constant pressure) is directly proportional to its temperature. As the temperature increases so does the volume that the gas occupies. The equations that describe this relationship are:
V ÷ T= constant
and
V1 ÷ T1 = V2 ÷ T2
It is possible to combine Boyle's law and Charles' law:
P1 × V1 ÷ T1 = P2 × V2 ÷ T2
The Ideal Gas Law includes the dependence of the volume on the number of gas moles, and an experimentally determined proportionality constant, called the Universal Gas Constant, which is designated by R.
The mathematical formulation of the law is written as follows:
(P × V) ÷ (n × T) = R
or in the more familiar form:
P × V = n × R × T
The value for the Universal Gas Constant, R, is 8.314472 J/mol*K (Joules per mole per degree Kelvin).
One important result of the Ideal Gas Law is that under conditions of constant pressure and temperature, one mole of any gas will always occupy the same volume.
The volume that one mole of an ideal gas occupies when held at a specific temperature and pressure is referred to as a "molar volume". For example, at one atmosphere, the molar volume of any ideal gas is 22.414 L/mol at 0 °C and 24.137 L/mol at 21°C. Most gases follow the Ideal Gas Law closely at atmospheric pressure and room temperature.
In this experiment you will determine the molar volume of hydrogen gas and compare it with the value predicted by the Ideal Gas Law. Hydrogen gas is produced in the following reaction between zinc and hydrochloric acid:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
One can calculate the molar volume of the gas at room temperature and pressure according to the equation:
Molar volume (L/mol) = (measured volume) ÷ (number of moles)
To complete this lab you will need to:
Calculate the number of moles of H2 produced. (Hint: For every mole of zinc used in the reaction, one mole of H2 is produced.)
Measure the volume of H2 gas produced.

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Part 1
The following are the givens from the experiment:
Initial Experiment with 10mL- 6 M HCl and 0.25g Zinc
Initial Teperature = 21.5C
Initial Pressure = 1.00atm
After adding 0.25g Zinc - The solution reacted
Pressure = 1.69atm
Temperature = 27.3C
Volume in Syrenge = 92.2mL

Part 2
Repeated Experiment with 10mL- 6 M HCl and 0.5g of Zinc
Initial Temperature = 21.5C
Initial Pressure = 1.00atm
After adding 0.5g Zinc -The solution reacted
Pressure = 2.41atm
Temperature =33.4C
Volume in Syrenge = 184.5mL

The following answers are what I have so far: Need to know if A,B,C,D,E, & F are correct for 1 & 2 if not please correct.

1. Record and calculate the following for the first half of the experiment:
a. Mass of zinc used (g) = 0.25g
b. Volume of 6M HCl used (mL) = 10mL
c. Molecular weight of zinc (g/mol) = 65.409 g/mol
d. Moles of zinc reacted = 0.25 g / 65.41 g/mol= 0.00382 moles
e. Moles of hydrogen produced = 0.00382 moles
f. Molar volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed as L/mol at X degrees C and a pressure of 1 atmosphere = 22.4 L/mole * ...

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