**Question-1[(a) (i)] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Given: When heated in air, magnesium ribbon reacts with oxygen to form magnesium oxide.**

**Discuss: Write a balanced equation for the reaction that occurs.**

**Answer/Explanation**

**Solution:**

The balanced chemical equation for the reaction between magnesium and oxygen can be written as:

$2 \mathrm{Mg}(\mathrm{s})+\mathrm{O} _2(\mathrm{~g}) \rightarrow 2 \mathrm{MgO}(\mathrm{s})$

This equation shows that two atoms of magnesium $(\mathrm{Mg})$ react with one molecule of oxygen gas $(\mathrm{O} _2)$ to form two molecules of magnesium oxide $(\mathrm{MgO})$ in solid state.

**Question-1[(a) (ii)] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Discuss: Identify a metal, in the same period as magnesium, that does not form a basic oxide.**

**Answer/Explanation**

**Solution:**

In the same period as magnesium (period 3), the metal aluminum $(\mathrm{Al})$ is another commonly known metal that does not form a basic oxide. Instead, aluminum reacts with oxygen to form aluminum oxide $(\mathrm{Al_2O_3})$, which is amphoteric and can behave as both an acid and a base.

**Question-1[(b) (i)] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Given: The reaction in (a)(i) was carried out in a crucible with a lid and the following data was recorded:**

**Mass of crucible and lid $=47.372 \pm 0.001 \mathrm{~g}$****Mass of crucible, lid and magnesium ribbon before heating $=53.726 \pm 0.001 \mathrm{~g}$****Mass of crucible, lid and product after heating $=56.941 \pm 0.001 \mathrm{~g}$**

**Calculate: the amount of magnesium, in mol, that was used.**

**Answer/Explanation**

**Solution:**

To calculate the amount of magnesium used in the reaction, we need to first determine the mass of magnesium used. We can do this by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and magnesium ribbon before heating:

mass of magnesium = (mass of crucible, lid, and Mg) – (mass of crucible and lid)

mass of magnesium $=(53.726 \pm 0.001 \mathrm{~g})-(47.372 \pm 0.001 \mathrm{~g})$

mass of magnesium $=6.354 \pm 0.002 \mathrm{~g}$

Next, we can use the molar mass of magnesium to convert the mass of magnesium used to moles of magnesium:

molar mass of $\mathrm{Mg}=24.31 \mathrm{~g} / \mathrm{mol}$

moles of $\mathrm{Mg}=\frac{(\mathrm{mass~ of~ Mg})}{(\mathrm{molar~ mass ~of~ Mg})}$

moles of $\mathrm{Mg}=\frac{(6.354 \pm 0.002 \mathrm{~g}) }{(24.31 \mathrm{~g} / \mathrm{mol})}$

moles of $\mathrm{Mg}=0.2613 \pm 0.0001 \mathrm{~mol}$

Therefore, the amount of magnesium used in the reaction is $0.2613 \pm 0.0001 \mathrm{~mol}$.

**Question-1[(b) (ii)] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Discuss: Determine the percentage uncertainty of the mass of product after heating.**

**Answer/Explanation**

**Solution:**

The mass of product can be calculated by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and product after heating:

mass of product $=$ (mass of crucible, lid, and product $)$ ( (mass of crucible and lid)

mass of product $=(56.941 \pm 0.001 \mathrm{~g})-(47.372 \pm 0.001 \mathrm{~g})$

mass of product $=9.569 \pm 0.002 \mathrm{~g}$

The absolute uncertainty in the mass of product is $\pm 0.002 \mathrm{~g}$.

The percentage uncertainty can be calculated using the formula:

percentage uncertainty $=(\frac{\mathrm{absolute~ uncertainty }}{\mathrm{ measured ~value}}) \times 100 \%$

percentage uncertainty $=(\frac{2 \times 0.001 \mathrm{~g}}{ 9.569 \mathrm{~g}}) \times 100 \%$

percentage uncertainty $=0.0209 \times 100 \%$

percentage uncertainty $=0.02 \%$

Therefore, the percentage uncertainty of the mass of product after heating is $0.02 \%$.

**Question-1[ (b) (iii)] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Discuss: ****Assume the reaction in (a)(i) is the only one occurring and it goes to completion, but some product has been lost from the crucible. Deduce the percentage yield of magnesium oxide in the crucible.**

**Answer/Explanation**

**Solution:**

To calculate the percentage yield of magnesium oxide, we need to know the theoretical yield and the actual yield. The theoretical yield is the amount of product that would be obtained if the reaction went to completion, and the actual yield is the amount of product that was actually obtained.

From the balanced equation in part (a)(i), we can see that two moles of magnesium react with one mole of oxygen to produce two moles of magnesium oxide. Therefore, the theoretical yield of magnesium oxide can be calculated as follows:

moles of $\mathrm{MgO}=2 \times$ moles of $\mathrm{Mg}$ used

moles of $\mathrm{MgO}=2 \times(0.2613 \pm 0.0001) \mathrm{mol}$

moles of $\mathrm{MgO}=0.5226 \pm 0.0002 \mathrm{~mol}$

To calculate the actual yield of magnesium oxide, we need to determine the mass of magnesium oxide obtained from the reaction. We can do this by subtracting the mass of the crucible and lid from the mass of the crucible, lid, and product after heating:

mass of $\mathrm{MgO}=$ (mass of crucible, lid, and product) – (mass of crucible and lid)

$$

\begin{aligned}

& \text { mass of } \mathrm{MgO}=(56.941 \pm 0.001 \mathrm{~g})-(47.372 \pm 0.001 \mathrm{~g}) \\

& \text { mass of } \mathrm{MgO}=9.569 \pm 0.002 \mathrm{~g}

\end{aligned}

$$

Next, we need to account for the possibility that some of the product was lost from the crucible. If we assume that the percentage loss is equal to the percentage uncertainty in the mass of the product, then we can calculate the corrected actual yield as follows:

corrected actual yield $=$ (mass of $\mathrm{MgO}) /(1$ – percentage loss $)$

percentage loss $=$ percentage uncertainty in mass of product $=0.02 \%$

corrected actual yield $=(9.569 \pm 0.002 \mathrm{~g}) /(1-0.02 \%)$

corrected actual yield $=(9.569 \pm 0.002 \mathrm{~g}) / 0.998$

corrected actual yield $=9.594 \pm 0.002 \mathrm{~g}$

Now we can calculate the percentage yield using the formula:

percentage yield $=($ corrected actual yield $/$ theoretical yield $) \times 100 \%$

percentage yield $=(9.594 \pm 0.002 \mathrm{~g}) /(0.5226 \pm 0.0002 \mathrm{~mol} \times 40.31 \mathrm{~g} / \mathrm{mol}) \times 100 \%$

percentage yield $=91.6 \pm 0.1 \%$

Therefore, the percentage yield of magnesium oxide in the crucible is $91.6 \pm 0.1 \%$.

**Question-1[(c) (i) ] :2022-may-Chemistry_paper_2__TZ1_HL**

**Topic:**

**Given: When magnesium is burnt in air, some of it reacts with nitrogen to form magnesium nitride according to the equation:**

$$

3 \mathrm{Mg}(\mathrm{s})+\mathrm{N}_2(\mathrm{~g}) \rightarrow \mathrm{Mg}_3 \mathrm{~N}_2(\mathrm{~s})

$$

**Evaluate: whether this, rather than the loss of product, could explain the yield found in (b)(iii).**

**Answer/Explanation**

**Solution:**

From part (b)(i), we know that the amount of magnesium used in the experiment is 0.2478 mol. Therefore, the maximum amount of magnesium nitride that could be formed is $(1 / 3) \mathrm{x}$ $0.2478 \mathrm{~mol}=0.0826 \mathrm{~mol}$

The molar mass of magnesium nitride is $100.95 \mathrm{~g} / \mathrm{mol}$. Therefore, the maximum mass of magnesium nitride that could be formed is:

$$

0.0826 \mathrm{~mol} \times 100.95 \mathrm{~g} / \mathrm{mol}=8.796 \mathrm{~g}

$$

This maximum mass of magnesium nitride is less than the mass of product obtained in the experiment $(9.569 \mathrm{~g})$. Therefore, the formation of magnesium nitride alone cannot explain the yield found in part (b)(iii).

In conclusion, the loss of product is the most likely explanation for the yield found in part (b)(iii).