question 3
grade 11 learners are conducting an experiment in a school laboratory and magnesium reacts with hydrochloric acid according to the following equation:
mg(s)+2hcl(aq)→mgcl₂(aq)+h₂(g)
3.1 define the term mole.
3.2 calculate the volume of hydrogen gas produced (at stp, molar volume = 22.4 dm³) when 4.8 g of magnesium reacts completely with excess hydrochloric acid.
3.3 state whether hydrochloric acid is the limiting reagent in this reaction. give a reason.
question 4
4.1 glucose (c₆h₁₂o₆) undergoes combustion as follows:
c₆h₁₂o₆(s)+6o₂(g)→6co₂(g)+6h₂o(l)
4.1.1 state avogadros law.
4.1.2 calculate the number of moles of oxygen required to completely burn 18 g of glucose.
4.2 a compound is - found to contain 2.40 g of sodium, 1.85 g of sulphur, and 3.70 g of oxygen. determine the empirical formula of the compound.
total: 30

Question

Accepted Answer

### 3.1
# Brief Explanations:
A mole is the amount of substance that contains as many elementary entities (such as atoms, molecules, or ions) as there are atoms in exactly 12 g of the carbon - 12 isotope.
# Answer:
A mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon - 12.

### 3.2
# Explanation:
## Step1: Calculate moles of magnesium
The molar mass of Mg is $M_{Mg}=24\ g/mol$. The number of moles of Mg, $n_{Mg}=\frac{m_{Mg}}{M_{Mg}}$, where $m_{Mg} = 4.8\ g$. So $n_{Mg}=\frac{4.8\ g}{24\ g/mol}=0.2\ mol$.
## Step2: Determine moles of hydrogen gas
From the balanced equation $Mg(s)+2HCl(aq)
ightarrow MgCl_{2}(aq)+H_{2}(g)$, the mole ratio of $Mg$ to $H_{2}$ is 1:1. So $n_{H_{2}} = n_{Mg}=0.2\ mol$.
## Step3: Calculate volume of hydrogen gas
At STP, the molar volume $V_{m}=22.4\ dm^{3}/mol$. The volume of $H_{2}$, $V_{H_{2}}=n_{H_{2}}	imes V_{m}$. So $V_{H_{2}}=0.2\ mol	imes22.4\ dm^{3}/mol = 4.48\ dm^{3}$.
# Answer:
$4.48\ dm^{3}$

### 3.3
# Brief Explanations:
The balanced equation shows that 1 mole of Mg reacts with 2 moles of HCl. Since it is stated that there is excess HCl, it means that HCl is not completely consumed in the reaction.
# Answer:
No. There is excess hydrochloric acid, so it is not completely consumed.

### 4.1.1
# Brief Explanations:
Avogadro's law states that equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.
# Answer:
Equal volumes of all gases, at the same temperature and pressure, contain the same number of molecules.

### 4.1.2
# Explanation:
## Step1: Calculate molar mass of glucose
The molar mass of glucose $C_{6}H_{12}O_{6}$, $M=(6	imes12 + 12	imes1+6	imes16)\ g/mol=180\ g/mol$.
## Step2: Calculate moles of glucose
The number of moles of glucose, $n_{glucose}=\frac{m_{glucose}}{M_{glucose}}$, where $m_{glucose}=18\ g$. So $n_{glucose}=\frac{18\ g}{180\ g/mol}=0.1\ mol$.
## Step3: Determine moles of oxygen
From the balanced equation $C_{6}H_{12}O_{6}(s)+6O_{2}(g)
ightarrow6CO_{2}(g)+6H_{2}O(l)$, the mole ratio of $C_{6}H_{12}O_{6}$ to $O_{2}$ is 1:6. So $n_{O_{2}}=6	imes n_{glucose}=6	imes0.1\ mol = 0.6\ mol$.
# Answer:
$0.6\ mol$

### 4.2
# Explanation:
## Step1: Calculate moles of each element
The molar mass of Na is $M_{Na}=23\ g/mol$, of S is $M_{S}=32\ g/mol$, and of O is $M_{O}=16\ g/mol$.
The number of moles of Na, $n_{Na}=\frac{m_{Na}}{M_{Na}}=\frac{2.40\ g}{23\ g/mol}\approx0.104\ mol$.
The number of moles of S, $n_{S}=\frac{m_{S}}{M_{S}}=\frac{1.85\ g}{32\ g/mol}\approx0.058\ mol$.
The number of moles of O, $n_{O}=\frac{m_{O}}{M_{O}}=\frac{3.70\ g}{16\ g/mol}\approx0.231\ mol$.
## Step2: Find the mole - ratio
Divide each number of moles by the smallest number of moles (0.058 mol).
For Na: $\frac{0.104\ mol}{0.058\ mol}\approx1.8$. Rounding to the nearest whole - number gives 2.
For S: $\frac{0.058\ mol}{0.058\ mol}=1$.
For O: $\frac{0.231\ mol}{0.058\ mol}\approx4$.
# Answer:
$Na_{2}SO_{4}$

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QUESTION IMAGE

Question

Explanation:

3.1

Step1: Calculate moles of magnesium

Step2: Determine moles of hydrogen gas

Step3: Calculate volume of hydrogen gas

Answer:

3.2