a) Using the mass of each piece of magnesium that you used, calculate the
number of moles of H2 expected in the reaction (use the reaction and
mole ratio). Show calculation for Trial 1 below.
The reaction we're seeing here is quite a simple one:
2HCl (aq) + Mg (s) -> MgCl2 (aq) + H2 (g)
As one can see the mole ratio of Mg to H2 is 1. Using this simple fact we can calculate the expected mole quantity of H2 after the reaction by calculating the molar quantity of magnesium in the initial state. For Trial 1, we measured out 0.035g of magnesium to react with. Using the formula
\begin{align*} \text{Mg (mol)} &= \frac{\text{Mg (g)}}{\text{Mg molar weight (g/mol)}} \\ &= \frac{0.035\text{g}}{24.3051\ \text{g/mol}} \\ &= 0.00144 \text{mol} \end{align*}We obtain 0.00144 mol as our answer for the amount of moles expected out of the reaction.
b) Calculate the volume change (\(\Delta V\)) from the reaction.
Show calculation for Trial 1 here.
For sealed vessels at standard temperature and pressure, we can use Boyle's Law to calculate the change in volume from a change in pressure. We use the formula to find this:
\begin{align*} P(\Delta V) &= V(\Delta P) \\ \Delta V &= \frac{V(\Delta P)}{P} \\ &= \frac{151 \text{mL} \cdot (1.063 \text{atm} - 0.8487 \text{atm})}{1 \text{atm}} \\ &= 38.128 mL \end{align*}With our answer in milliliters, we can divide it by 1000 to get it in liters: 0.038128 L.
c) Calculate the volume of one mole of hydrogen gas (\(\Delta V / n\)).
Show calculation for Trial 1 here
Now with our calculation for the volume of the amount of hydrogen we produced, we can simply divide our volume by our molar quantity of H2 to find the volume for a whole mole of H2 under RTP. Like so:
\begin{align*} \text{Volume} &= \frac{\Delta V}{n} \\ &= \frac{0.038128 \text{L}}{0.00144 \text{mol}} \\ &= 26.477 \frac{\text{L}}{\text{mol}} \end{align*}
d) Calculate the number of moles of hydrogen gas produced (\(\Delta n\))
Show calculations for Trial 1 here.
This is one of the more complicated parts of the calcculations, and it leverages the ideal gas law: \(PV = nRT\). Where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of moles, \(R\) is the Boltzmann constant, which is around \(8.314462 \frac{\text{J}}{\text{K mol}}\), and \(T\) the temperature. Rearranging this formula, we can find the amount of moles we produced:
\begin{align*} PV &= nRT \\ n &= \frac{PV}{RT} \\ &= \frac{0.2142 \text{atm} \cdot 0.0151 \text{L}}{8.314462 \frac{\text{J}} {\text{K mol}} \cdot (28.2 \text{degC} + 273.15 \text{degC}) \text{K}} \\ &= 0.001309 \text{mol} \end{align*}
e) Convert the molar volume at room temperature and pressure (step c) to STP.
Show Trial 1 here.
Using our molar volume at RTP and leveraging the combined gas law we can find this:
\begin{align*} \frac{PV}{T}_{\text{RTP}} &= \frac{PV}{T}_{\text{STP}} \\ V_{\text{STP}} &= \frac{P_{\text{RTP}} V_{\text{RTP}} T_{\text{STP}}} {T_{\text{RTP}} P_\text{STP}} \\ &= \frac{0.2142 \text{atm} \cdot 26.477 \frac{\text{L}}{\text{mol}} \cdot 273.15 \text{K}} {(28.2 \text{degC} + 273.15 \text{K})\text{K} \cdot 1 \text{atm}} \\ &= 20.3685 \frac{\text{L}}{\text{mol}} \end{align*}| TABLE: Molar Volume of a Gas | Trial 1 | Trial 2 | Trial 3 |
|---|---|---|---|
| Atomic Weight of Magnesium (g/mol) | 24.3051 | 24.3051 | 24.3051 |
| Volume of Flask, \(V\), (mL) | 151 | 151 | 151 |
| Mass of Mg (g) | 0.035 | 0.045 | 0.034 |
| Initial pressure, \(P_i\) (atm) | 0.8487 | 0.8586 | 0.8345 |
| Final pressure, \(P_f\) (atm) | 1.063 | 1.215 | 1.104 |
| Room Temperature, \(t\) (degC) | 28.2 | 27.7 | 26.6 |
| Pressure Change, \(\Delta P\) (\(P_f\)-\(P_i\)) (atm) | 0.2142 | 0.3564 | 0.2695 |
| Moles Mg (mol) | 0.00144002 | 0.00185146 | 0.00139888 |
| Volume of H2 produced, \(\Delta V\) (L) | 0.038128 | 0.062679 | 0.048765 |
| Calculated Molar Volume H2, \(\frac{V}{n}\) (L/mol) | 26.4773 | 33.8539 | 34.8600 |
| Moles H2 produced in experiment, \(\Delta n\) (mol) | 0.001309 | 0.002180 | 0.001654 |
| Calculated molar volume (L/mol) | 26.4773 | 33.8539 | 34.8600 |
| Calculated molar volume at STP (L/mol) | 20.3685 | 26.3907 | 26.5092 |
| Expected molar volume at STP (L) | 22.414 | 22.414 | 22.414 |
| % Error | 9.1261 | -17.7420 | -18.2706 |
| Average % Error | -8.9621 |
1. Compare your calculated molar volume, at STP, with the accepted molar volume
of an ideal gas at STP, 22.4 L/mol. Suggest possible sources of experimental
error.
Instruments used for this experiment were of varying quality, our way of measuring volume of the erlenmeyer flask was suboptimal, haste of operators.
2. Compare the moles of H2 produced and expected. Comment on their
similarities/differences.
For Trial 1, our moles of H2 produced was lower than expected. This is common, as the reaction is never perfectly completed. However, for Trials 2 and 3, our produced amount of H2 was higher than expected. This is quite odd. I'm not quite sure where this discrepancy could have occurred, unless our reader was faulty.
3. If this experiment had been performed using a reaction that produced oxygen
gas instead of hydrogen gas do you think the calculated value for the molar
volume of a gas would be the same or different? Explain your reasoning.
This would probably not be different. Assuming, of course, that the reaction that produces oxygen gas does not produce other gases. The trouble is, of course, it's difficult to find a reaction that just produces oxygen gas that doesn't include something else that would fudge up the numbers.