In an eudiometer 20cm^3 of a gaseous hydrocarbon A is mixed with 150 cm^3 of oxygen. The combustion of A is initiated by an electric spark. After cooling and condensing the vapor the volume of the gaseous mixture is 110cm^3 of which 80cm^3 causes turbidity of limewater. The remaining volume of the gas makes a glowing splint to catch fire.
Write the equation of the combustion of A. Determine the molecular formula of A.
The action of hydrogen chloride on A gives one single compound B. Write the equation of its formation and write its name.
THANKS.
Solve this parr
To determine the equation of the combustion of hydrocarbon A and its molecular formula, we need to analyze the given information.
Step 1: Determine the reactants and products in the combustion reaction:
The reactants are hydrocarbon A (CnHm) and oxygen (O2).
The products of the combustion reaction are carbon dioxide (CO2) and water (H2O).
Therefore, the balanced equation for the combustion of hydrocarbon A is:
CnHm + (n + m/4)O2 → nCO2 + (m/2)H2O
Step 2: Use the given information to determine the values of n and m.
From the information provided, we know that the initial volume of hydrocarbon A is 20 cm^3, and after combustion, the resulting volume of the gaseous mixture is 110 cm^3.
The volume of the mixture causing turbidity in limewater (CO2) is 80 cm^3.
The remaining volume of the gas (unreacted oxygen + water vapor) that ignites a glowing splint is the difference: 110 cm^3 - 80 cm^3 = 30 cm^3.
From this information, we can conclude that the volume of CO2 produced is equal to the volume of hydrocarbon A consumed. Therefore, the mole ratio of CO2 to hydrocarbon A is 1:1.
The molecular formula of hydrocarbon A can be determined by dividing the volume of hydrocarbon A by the volume of 1 mole (22.4 L) at standard temperature and pressure (STP). In this case, the volume of hydrocarbon A is 20 cm^3.
20 cm^3 ÷ 22.4 L = 0.00089286 moles = 0.00089286n
Since the mole ratio of hydrocarbon A to CO2 is 1:1, the mole ratio of hydrocarbon atoms to carbon atoms in the hydrocarbon A is also 1:1.
Therefore, the molecular formula of hydrocarbon A can be written as C1H?
Step 3: Determine the molecular formula of hydrocarbon A.
To find the value of m in the hydrocarbon's molecular formula, we need to consider the volume of water vapor formed. From the given information, the volume of water produced is half the volume of hydrocarbon A.
The volume of water vapor formed is (1/2) × 20 cm^3 = 10 cm^3.
The mole ratio of water vapor to hydrocarbon A is 0.447:1.
0.447 moles of water = (0.447 × 18 g) ÷ 1000 g = 0.008046 g of water
0.008046 g of water is equivalent to (0.008046 g ÷ 18 g/mol) = 0.000447 moles of water.
The mole ratio of water vapor to moles of hydrocarbon A is approximately 1:1. Therefore, m = 2.
Hence, the molecular formula of hydrocarbon A is CH2.
Step 4: Determine compound B formed from the reaction with hydrogen chloride.
When hydrocarbon A reacts with hydrogen chloride (HCl), a single compound B is formed. To determine the equation of its formation, we need to consider the reaction between chlorination and the hydrocarbon.
The equation for the formation of compound B can be written as:
CH2 + HCl → CH3Cl
The name of compound B is chloromethane.
To summarize:
- The equation of the combustion of hydrocarbon A is CnHm + (n + m/4)O2 → nCO2 + (m/2)H2O.
- The molecular formula of hydrocarbon A is CH2.
- The equation of the formation of compound B (chloromethane) is CH2 + HCl → CH3Cl.
To find the equation of the combustion of hydrocarbon A and determine its molecular formula, we need to understand the stoichiometry of the reaction.
Let's start by analyzing the given information:
1. Initial volume of hydrocarbon A = 20 cm^3
2. Volume of oxygen = 150 cm^3
3. Volume of the gaseous mixture after combustion = 110 cm^3
4. Volume of the gas causing turbidity in limewater = 80 cm^3
5. Volume of the gas that makes a glowing splint catch fire = Remaining volume
Now, let's determine the equation of the combustion of hydrocarbon A:
1. The hydrocarbon A combusts with oxygen to produce carbon dioxide (CO2) and water (H2O) as the combustion products.
2. The balanced equation for this reaction can be written as:
CxHy + (x+y/4)O2 → xCO2 + (y/2)H2O
To find the values of x and y, we need to equate the volumes of the reactants and products using the ideal gas law. First, let's calculate the volume of oxygen required for complete combustion:
20 cm^3 of A + Volume of Oxygen = 110 cm^3 of Products
Volume of Oxygen = 110 cm^3 - 20 cm^3
Volume of Oxygen = 90 cm^3
Now, we can equate the volumes of oxygen for the combustion:
150 cm^3 of Oxygen = 90 cm^3 of Oxygen
Therefore, the ratio of Oxygen:Hydrocarbon A is 90:20 or 9:2.
Now, let's determine the values of x and y in the balanced equation:
From the ratio of Oxygen:Hydrocarbon A, we can see that x should be 9 and y should be 2.
The equation for the combustion of A is therefore:
C2H4 + 9O2 → 2CO2 + 2H2O
To determine the molecular formula of hydrocarbon A, we need to find the empirical formula first. The empirical formula shows the simplest whole number ratio of elements present.
From the balanced equation, we can see that there are two carbons (C) and four hydrogens (H) in the empirical formula.
Thus, the empirical formula of hydrocarbon A is CH2.
To determine the molecular formula, we need to know the molecular mass of hydrocarbon A. Without this information, we cannot determine the molecular formula. Please provide the molecular mass of hydrocarbon A to proceed.
Regarding the formation of compound B by the action of hydrogen chloride (HCl) on A, we would need additional information such as the reaction conditions or the products formed to write the equation and name the compound. Please provide the necessary information so that we can assist you further.