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Mg+HCl Lab report about rate of reaction

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Chemistry Lab Report from the 19th December 2011 Introduction In the chemistry lab on Monday the 19th December 2011, we were supposed to recognize that different salts can change the colour(s) of a flame. We gave with each salt a little ethanol (just as much that the floor of the petri dish was covered and maybe a little bit more) to a petri dish, mixed it well with the spoon spatula, ignited the mixture and than noted the colour(s) of the flame and how the flame changed. While the mixture was burning, we took a video from the event. Every time…
Mg+HCl Lab report about rate of reaction Table 1: Table 2:

all the calculation is shown on the table

Table 3: molar volume = volume gas/n = 0.0048L / 0.00022mol = 21.818l/mol Table 4:

uncertainty = 0.01 / 0.02 × 100% = 50%

average uncertainty = (50% + 50% + 50%) / 3=50%

Table 5: average molar volume = (21.818 + 21.411 + 21.667) L/mol / 3 =21.6320 L/mol

uncertainties from repeats = maximum – minimum = 21.818 - 21.411 = 0.407

percentage of uncertainties = 0.407 / 21.6320 × 100% = 1.881%

Table 1 shows all the raw data that collecting in the lab. The last column: p×V is calculated for the following calculation. During this experiment, there’s obvious outlier which happens in Table 1 and causes the following calculations all wrong.

Compare to group of 3cm and 5cm Mg, the group of 4cm Mg has wrong data for the volume of H2 measured, which steady in about 10mL (0.01L). But have disproportionate mass and pressure. For group of 5cm Mg, although the errors are not so obvious, the volumes of H2 may have some problems in measuring since the volumes are relative low compare to other groups and finally cause the average value of molar volume of H2 is not close to the theoretical value that about 22.7mol/L.

For Table 2, R is a constant that using in the calculation directly. T is the room temperature in form of Kelvin. p×V uses the previous calculated data. According to the formula pV=nRT, n= . To make sure the following calculations are as previse as possible, I make the n has 6 significance digit, which one more time than the usual one.

For Table 3, the significance digits are all more than common ones. I use 5 significance digit here since the final ones are relative high which means that less significance digit can also shows the difference.

In Table 4, the total percentage uncertainty is 32.458%, which is relative high and shows the measure during the experiment is not accurate. However, since 32.458% is the adds of all separate % errors: 28.24%, 4.21% and 0.008254%, and it can be noticed that the error is caused by 28.24%, which is the mass of Mg measured.

The high inaccuracy of the mass of Mg is generally caused by the small difference between the mass of Mg and the uncertainty, which is in group of 1, 2 and 3. On the other hand, the other uncertainties, like 4.21% and 0.008254% are really small, which means that hig.....[read full text]

1≈ †+∞ 7∋+†∞ 5, †+∞ ⊥∞+≤∞≈†∋⊥∞ ∞++++ †++ †+∞ 6 ∋∋≈;⊥∞†∋†;+≈≈ ∋+∞ 1.881%, 2.716%, 6.394%, 3.515% ∋≈⊇ 0.166%. 7+∞ ∞++++≈ +∋=∞ ≈+∋∞ ⊇;††∞+∞≈≤∞ +∞†≠∞∞≈ ∞∋≤+ +†+∞+ +∞† ∋†† +† †+∞∋ ∋+∞ +∞†∋†;=∞ †+≠. +†+∞+, ≈+∋∞ ∋+∞ +∞†∋†;=∞†+ ≈;⊥≈;†;≤∋≈† ≠+;†∞ ≈+∋∞ ∋+∞ ⊥∞;†∞ ;≈≈;⊥≈;†;≤∋≈† ≠;†+ ∋ ≈∋∋†† ≈∞∋+∞+ †;∂∞ 10.49%.

8∞∞ †+ †+∞ +∞††;≈∞+, †+∞ ∋=∞+∋⊥∞ ∋+†∋+ =+†∞∋∞ ;≈ +≈†+ 20.452 7/∋+†. 7+∞ +=∞+∋†† ∋=∞+∋⊥∞ ∋+†∋+ =+†∞∋∞ ∞≠≤∞⊥† †+∞ +∞††;∞+ ;≈ 20.6437/∋+†, ≠+;≤+ ;≈ +∞†∋†;=∞†+ ≤†+≈∞ †+ †+∞ †+∞++∞†;≤∋† =∋†∞∞ 22.77/∋+†. 7+∞ +∞∋≈+≈ †+∋† †+∞ †;≈∋† +≈∞ ;≈ ≈†;†† †+≠∞+ †+∋≈ †+∞ †+∞++∞†;≤∋† ;≈ †+∋† †+∞ ⊥++∞⊥ +† 5 ≤∋ 4⊥ ∋∋+ ∋†≈+ +∋=∞ ≈+∋∞ ∞++++≈ ;≈ ∋∞∋≈∞+;≈⊥ †+∞ =+†∞∋∞ +† 82, +∞† †+∞ ∞++++ ;≈ ≈+† ≈+ ++=;+∞≈ ≤+∋⊥∋+∞ †+ †+∞ ⊇∋†∋ +† ⊥++∞⊥ +† 4 ≤∋ 4⊥.

1 ≤+∞†⊇ ≈+† ∞≈≈∞+∞ †+∋† †+∞+∞ ∋+∞ ⊥+++†∞∋≈ ;≈ †+;≈ ≈†∋⊥∞.

4≤≤++⊇;≈⊥ †+ †+∞ †++∋∞†∋: 0∞+≤∞≈†∋⊥∞ 9++++ = [(7+∞∞ 2∋†∞∞ - 9≠⊥∞+;∋∞≈†∋† 2∋†∞∞) / 7+∞∞ 2∋†∞∞] ≠ 100%, ;≈ †+;≈ ;≈=∞≈†;⊥∋†;+≈, ⊥∞+≤∞≈†∋⊥∞ ∞++++ = [(22.7 - 20.643) / 22.7] ≠ 100% ≈ 9.06%, ≠+;≤+ ∋∞∋≈≈ †+∞ ∞++++ ;≈ +∞†∋†;=∞†+ ≈∋∋††. 1† ≤∋≈ +∞ ≤+≈≈;⊇∞+∞⊇ ∋≈ ∋ +∞†∋†;=∞†+ ;≈≈;⊥≈;†;≤∋≈† ∞++++ ∋≈⊇ †+∞ +∞†≤+∋∞ ;≈ ⊥∞≈∞+∋††+ ≤+++∞≤†, ∋≈⊇ +∞†∋†;=∞†+ ∋≤≤∞+∋†∞.

1≈ †+;≈ ;≈=∞≈†;⊥∋†;+≈, †+∞ ∋;∋ ;≈ †+ ⊇∞†∞+∋;≈∞ †+∞ ∋+†∋+ =+†∞∋∞ +† 82. 4≤≤++⊇;≈⊥ †+ 4=+⊥∋⊇++’≈ ++⊥+†+∞≈;≈ †+∋† ∞⊥∞∋† =+†∞∋∞≈ +† ∋†† ⊥∋≈∞≈ ≤+≈†∋;≈ ∞⊥∞∋† ≈∞∋+∞+≈ +† ∋+†∞≤∞†∞≈ ∞≈⊇∞+ †+∞ ≈∋∋∞ ≤+≈⊇;†;+≈≈ +† †∞∋⊥∞+∋†∞+∞ ∋≈⊇ ⊥+∞≈≈∞+∞. 3+, ;† 1 ≤∋≈ ⊇∞†∞+∋;≈∞ †+∞ ∋∋≈≈ +† 4⊥, 1 ≤∋≈ ⊥∞† †+∞ ≈∞∋+∞+ +† ∋+†∞≈ +† 4∋⊥≈∞≈;∞∋ ∋≈⊇ †+∞ ≈∞∋+∞+ +† ∋+†∞≈ +† 82.

7+∞ ≤+∞∋;≤∋† †++∋∞†∋ ;≈:

4⊥ + 280† 4⊥0†2+ 82

6++∋ †+∞ ≤∋†≤∞†∋†;+≈, ≠∞ ≤+∞†⊇ ∂≈+≠ †+∋† †+∞ ∞≠⊥∞+;∋∞≈†∋† =∋†∞∞ +† †+∞ ∋+†∋+ =+†∞∋∞ +† 82 ;≈ 20.643 7/∋+†, ≠+;≤+ ;≈ +∞†∋†;=∞†+ ≈∋∋††∞+ ≤+∋⊥∋+∞ †+ †+∞ †+∞++∞†;≤∋† =∋†∞∞ 22.7 7/∋+†. 1† 1 ⊇∞†∞†∞ +≈∞ ≈∞≈⊥∞≤†∞⊇ ⊥++∞⊥ +† ⊇∋†∋ ∋⊥∋;≈, †+∞ ⊇∋†∋ ≠;†† ≤†+≈∞+ †+ †+∞ †+∞++∞†;≤∋† =∋†∞∞.

7+∞ ∋=∞+∋⊥∞ ≈∞∋ ∞≈≤∞+†∋;≈†+ ;≈ †+;≈ ∞≠⊥∞+;∋∞≈† ;≈ +≈†+ 10.819%, ≠+;≤+ ∋∞∋≈≈ †+∋† †+∞ ⊥∞≈∞+∋† +∞†≤+∋∞ ;≈ ∋≤≤∞+∋†∞.

In this investigation, the Molar Volume of H2 is measured by a different method than the Task sheet provided. I determine the mass of Mg, the volume and the pressure of H2 and the temperature, which are necessary for the ideal gas law. All the necessary data are listed in Table 1 and 2.

In Table 3, the molar volume of hydrogen gas is obtained by converting the ideal gas law into n= . The average molar volume of H2 is 20.643 L/mol and the differences might be caused by the errors and uncertainties occurred in the experiment like the escaping of H2 during the procedure.

The systematic error of the investigation is 32.458% which is significant, which means that there needs to be improvements for the setting of the investigation. Since this great error is generally caused by the mass of Mg, we can use heavier Mg to redu.....

6++ †+∞ +†+∞+ †≠+, †+∞+ ∋+∞ +∞†∋†;=∞ ≈∋∋†† ∋≈⊇ ∋∋+ ≈+† ∋††∞≤† †+∞ +=∞+∋†† ∞≠⊥∞+;∋∞≈†.

0=∞+∋††, †+;≈ †∋+ ;≈ ∋ ≈∞≤≤∞≈≈†∞† ;≈=∞≈†;⊥∋†;+≈ ∋++∞† †;≈⊇;≈⊥ †+∞ ∋+†∋+ =+†∞∋∞ +† ⊥∋≈ (≈⊥∞≤;†;≤ †++ 82) ∋† 370. 4≈⊇ †+∞ ∞≠⊥∞+;∋∞≈†∋† =∋†∞∞ ≠∞ ⊥+† ;≈ +∞†∋†;=∞ ≤†+≈∞ †+ †+∞ †+∞++∞†;≤∋† =∋†∞∞ +† 22.77/∋+†.

4†† †+∞ ≈†∞⊥≈ +† †+∞ ∞≠⊥∞+;∋∞≈†∞+≈ ∋+∞ †+∞ ≈∋∋∞, ≠+;≤+ ∋∞∋≈≈ †+∋† †+∞ ≠++†∞ ≤†∋≈≈ ≠;†† ≈+† +∋=∞ ∋;≈†∋∂∞≈ ;≈ ≠++≈⊥ +⊥∞+∋†;+≈≈ †+∋† †∞∋⊇ †+ ≠++≈⊥ ⊇∋†∋. 6++ †+;≈ ;≈=∞≈†;⊥∋†;+≈, †+∞+∞’+∞ 5 ∋∋≈;⊥∞†∋†;+≈≈ ≠;†+ ∋≈ ;≈≤+∞∋≈;≈⊥ +† 1.0≤∋ ∞=∞++ ∋∋≈;⊥∞†∋†;+≈ ∋≈⊇ 3 †+;∋†≈, ≠+;≤+ ∋∞∋≈≈ ∋ ∋++∞ ∋≤≤∞+∋†∞ ⊇∋†∋ †∋+†∞ †+∋† +∞†⊥ ∞≈ +∋=∞ ∋ ⊥+∞≤;≈∞ +∞≈∞††.

1≈ †+∞ ⊥++≤∞≈≈ +† †+∞ †∋+, †+∞+∞’≈ +≈∞ ≈†∞⊥ ≠+;≤+ ;≈ ‘⊇++⊥ †+∞ 4⊥ ;≈†+ †+∞ ≤+≈;≤∋† ††∋≈∂ ∋≈⊇ ⊥∞;≤∂†+ ≤+=∞+ ∞⊥ †+∞ ≠++†∞ ++ †+∞ +∞++∞+ ≈†+⊥⊥∞+’, ++≠∞=∞+, †+;≈ ≈†∞⊥ ∋∋+ ≤∋∞≈∞ †+∞ ;∋⊥+∞≤;≈∞ +† †+∞ ⊇∋†∋ ≤+††∞≤†∞⊇ ≈;≈≤∞ ;† ≠;†† †∋∂∞ ∋ ⊥∞+;+⊇ +† †;∋∞ †+ ≤†+.....

4≈⊇ †+∞ 4⊥ ≠;†† +∞∋≤† ≠;†+ 80† ≠+∞≈ †+∞+ †+∞≤+ ∞∋≤+ +†+∞+. 7+;≈ ∋∞∋≈≈ †+∋† †+∞+∞ ∋∋+ +∞ ∋∋+∞≈† +† 82 ∞≈≤∋⊥∞ +∞†++∞ ≠∞ ≤+=∞+ †+∞ +∞++∞+ ≈†+⊥⊥∞+. 6++ †+∞ ⊥+≈≈;+†∞ ;∋⊥++=∞∋∞≈†≈, ≈∞≠† †;∋∞ ≠+;†∞ ⊇+;≈⊥ †+∞ †∋+, ≠∞ ∋∋+ ⊥∞† †+∞ 4⊥ ;≈ †+∞ ≤+≈;≤∋† ††∋≈∂ †;+≈††+ ∋≈⊇ ∞≈∞ ≈++;≈⊥∞ †+ ∋⊇⊇ 80† ;≈†+ †+∞ ≤+≈†∋;≈∞+.

1≈ †+;≈ ≠∋+, ≠∞ ≤+∞†⊇ ∋∋∂∞ ≈∞+∞ †+∋† †+∞ 82 ∋∋+ ≈+† ∞≈≤∋⊥∞ ∋≈⊇ ≤∋∞≈∞ ∞++++≈ ;≈ †+∞ ⊇∋†∋ ≤+††∞≤†;≈⊥.

4†≈+, ≠+∞≈ ≠∞ ⊇;⊇ †+∞ ∞≠⊥∞+;∋∞≈†, 1 †+∞≈⊇ †+∋† ≠+∞≈ ≠∞ ∋∞∋≈∞+∞⊇ †+∞ =+†∞∋∞ +† 82, ≠∞ ≈∞∞⊇ †+ ⊥∞≈+ †+∞ ⊥;≈†+≈ +† †+∞ ≈++;≈⊥∞ †++ ∋ †;†††∞ +;†, +†+∞+≠;≈∞, †+∞ ≈++;≈⊥∞ ≠;†† ≈+† ⊥+ ∞⊥ ∋≈+∋++∞. 7+;≈ ∋∋+ ≤∋∞≈∞ †≠+ ∂;≈⊇≈ +† ≈;†∞∋†;+≈. 1† †+∞ =+†∞∋∞ +† 4⊥ ;≈ ≈+† ≈+ ∋∞≤+, †;∂∞ †++ ⊥++∞⊥ +† 1 ≤∋ 4⊥.

When we push the piton, it may go beyond the real volume of H2. If the volume of Mg is close to the range of the syringe, this operation will cause the volume goes beyond the range and the data will be wrong for the calculation. To improve this situation, I think the easiest method is using another equipment that can detect the volume of H2 to replace the syringe.

This could prevent some of the errors during the experiment.

During this experiment, the key method is using the mass, volume and pressure of a kind of gas to detect its molar volume and prove the ideal gas volume. By using the similar method, we could do some extension. Like using Zn to replace Mg. The chemical formula is Zn+ 2HCl ZnCl2 + H2.

This experiment could also be done as a method to comparing the activity between Zn and Mg by comparing the time that the .....