Question

physical science - control test 6 nwseptember 2025
question 3
3.1 compound q (cxhy) reacts with oxygen according to the balanced equation:
p cxhy + 13 o2→8 co2 + 10 h2o
the molar mass of compound q is 58 g·mol⁻¹
3.1.1 define empirical formula.
3.1.2 use the principle of conservation of mass and determine the value p.
the percentage composition of compound q is:

carbon	hydrogen
82.76%	17.24%

3.1.3 determine the molecular formula of compound q.
3.2 5 g of sodium carbonate (na2co3) reacts with 250 cm³ of hydrochloric acid (hcl).
na2co3(s) + 2 hcl(aq)→2 nacl(aq)+ h2o(l)+ co2(g)
the percentage of hydrochloric acid (hcl) that reacted with sodium carbonate (na2co3) is 76%.
3.2.1 define the term limiting reagent.
calculate the:
3.2.2 amount of hydrochloric acid that reacted with sodium carbonate.
3.2.3 volume of carbon dioxide that was produced. take the molar volume at room temperature as 24.43 dm³.
3.2.4 the hydrochloric acid that was used in the reaction was obtained by diluting 100 cm³ hcl to 250 cm³ hydrochloric acid (hcl) solution.
calculate the concentration of the concentrated hydrochloric acid (hcl).

Explanation:

3.1.1

The empirical formula of a compound is the simplest whole - number ratio of atoms of the elements present in the compound.

3.1.2

Step1: Balance carbon atoms

From the products, there are 8 carbon atoms in $8CO_2$. In the reactant side, compound $Q$ is $C_xH_y$. So, from the conservation of carbon atoms, $P\times x = 8$.

Step2: Balance hydrogen atoms

There are 20 hydrogen atoms in $10H_2O$. So, $P\times y=20$.

Step3: Consider the molar - mass information later

We know that the molar mass of $Q$ is $58\ g/mol$. Let's first find the empirical formula and then the molecular formula. For now, balancing the oxygen atoms: on the product side, there are $8\times2 + 10\times1=26$ oxygen atoms. On the reactant side, $13O_2$ provides 26 oxygen atoms. Since the equation is balanced for oxygen already, and considering carbon and hydrogen balance, if we assume the empirical formula of $Q$ is $C_xH_y$ and from carbon and hydrogen balance in the reaction $P\times x = 8$ and $P\times y = 20$. The ratio of $x:y = 8:20=2:5$. So the empirical formula of $Q$ is $C_2H_5$. The empirical - formula mass $EFM=(2\times12)+(5\times1)=29\ g/mol$. Since the molar mass of $Q$ is $58\ g/mol$, $n=\frac{M}{EFM}=\frac{58}{29}=2$. So the molecular formula of $Q$ is $C_4H_{10}$ and $P = 2$.

3.1.3

Step1: Calculate moles of carbon and hydrogen from percentage composition

Let's assume we have a 100 - g sample of compound $Q$.
Moles of carbon, $n_{C}=\frac{82.76\ g}{12\ g/mol}=6.897\ mol$
Moles of hydrogen, $n_{H}=\frac{17.24\ g}{1\ g/mol}=17.24\ mol$

Step2: Find the ratio of moles

The ratio of $n_{C}:n_{H}=\frac{6.897}{6.897}:\frac{17.24}{6.897}\approx1:2.5 = 2:5$ (multiplying by 2 to get whole - numbers). So the empirical formula is $C_2H_5$.

Step3: Determine the molecular formula

The empirical - formula mass $EFM=(2\times12)+(5\times1)=29\ g/mol$. Given the molar mass of $Q$ is $58\ g/mol$. Let $n$ be the ratio of molar mass to empirical - formula mass, $n=\frac{58}{29}=2$. So the molecular formula is $C_4H_{10}$.

3.2.1

The limiting reagent is the reactant that is completely consumed in a chemical reaction and limits the amount of product that can be formed.

3.2.2

Step1: Calculate moles of $Na_2CO_3$

The molar mass of $Na_2CO_3$ is $M=(2\times23)+12+(3\times16)=106\ g/mol$. Moles of $Na_2CO_3$, $n_{Na_2CO_3}=\frac{5\ g}{106\ g/mol}=0.0472\ mol$

Step2: Use the stoichiometric ratio

From the balanced equation $Na_2CO_3(s)+2HCl(aq)\to2NaCl(aq)+H_2O(l)+CO_2(g)$, the mole ratio of $Na_2CO_3:HCl = 1:2$. Moles of $HCl$ that react with $Na_2CO_3$ is $n_{HCl}=2\times n_{Na_2CO_3}=2\times0.0472\ mol = 0.0944\ mol$

3.2.3

Step1: Use the stoichiometric ratio for $CO_2$

From the balanced equation, the mole ratio of $Na_2CO_3:CO_2 = 1:1$. Moles of $CO_2$ produced, $n_{CO_2}=n_{Na_2CO_3}=0.0472\ mol$

Step2: Calculate volume of $CO_2$

Given the molar volume at room temperature $V_m = 24.43\ dm^3/mol$. Volume of $CO_2$, $V=n_{CO_2}\times V_m=0.0472\ mol\times24.43\ dm^3/mol = 1.153\ dm^3$

3.2.4

Step1: Calculate moles of $HCl$ in the diluted solution

We know that the amount of $HCl$ that reacted is $n_{HCl}=0.0944\ mol$, and this is 76% of the $HCl$ in the diluted solution. Let the moles of $HCl$ in the diluted solution be $n_{total\ HCl}$. Then $0.76\times n_{total\ HCl}=0.0944\ mol$, so $n_{total\ HCl}=\frac{0.0944\ mol}{0.76}=0.1242\ mol$

Step2: Calculate concentration of the concentrated $HCl$

The volume of the diluted solution is $V_{diluted}=250\ cm^3 = 0.25\ dm^3$. Using the dilution formula $n = C\times V$. For the concentrated solution, $V_{concentrated}=100\ cm^3=0.1\ dm^3$. Concentration of the concentrated $HCl$, $C=\frac{n_{total\ HCl}}{V_{concentrated}}=\frac{0.1242\ mol}{0.1\ dm^3}=1.242\ mol/dm^3$

Answer:

3.1.1: The simplest whole - number ratio of atoms of elements in a compound.
3.1.2: $P = 2$
3.1.3: $C_4H_{10}$
3.2.1: The reactant that is completely consumed in a chemical reaction and limits the amount of product formed.
3.2.2: $0.0944\ mol$
3.2.3: $1.153\ dm^3$
3.2.4: $1.242\ mol/dm^3$