A hydrocarbon has mass ratio of C and H in 12 : 1.
Each molecule of hydrocarbon has 2 carbon atoms.
Calculate mass of \( CO_2 \) (in gm) produced, when 3.38 gm hydrocarbon undergoes combustion.

Question

Correct statement about product X and Y

\(\text{(a) They can be differentiated by NaHCO}_3 \text{ test.} \)
\(\text{(b) They both gives 2,4-DNP test.} \\ \text{(c) They both have same molecular mass.} \\ \text{(d) They both react with same rate with HCN.} \)

Answer 1

Step 1: Understanding the hydrocarbon.
Let the molecular formula of the hydrocarbon be \( C_x H_y \). Given that the mass ratio of C and H is 12 : 1, the molar mass ratio of C and H is 12:1. Each molecule contains 2 carbon atoms, so the molecular formula of the hydrocarbon is \( C_2 H_4 \) (ethene).

Step 2: Calculate the molar mass of the hydrocarbon.
The molar mass of \( C_2 H_4 \) is calculated as:
\[ M = (2 \times 12) + (4 \times 1) = 24 + 4 = 28 \, \text{gm/mol} \]
Step 3: Find the number of moles of hydrocarbon.
The number of moles of hydrocarbon in 3.38 gm is given by: \[ \text{moles of hydrocarbon} = \frac{3.38}{28} = 0.120 \, \text{mol} \]
Step 4: Calculate the number of moles of \( \text{CO}_2 \) produced.
From the balanced combustion reaction of \( C_2 H_4 \): \[ C_2 H_4 + 3 O_2 \rightarrow 2 CO_2 + 2 H_2 O \] We see that 1 mole of \( C_2 H_4 \) produces 2 moles of \( CO_2 \). Therefore, 0.120 moles of \( C_2 H_4 \) will produce: \[ 2 \times 0.120 = 0.240 \, \text{mol of } CO_2 \]
Step 5: Calculate the mass of \( \text{CO}_2 \) produced.
The molar mass of \( CO_2 \) is: \[ M_{\text{CO}_2} = (12 \times 1) + (16 \times 2) = 44 \, \text{gm/mol} \] The mass of \( CO_2 \) produced is: \[ \text{mass of } CO_2 = 0.240 \times 44 = 10.56 \, \text{gm} \]
Step 6: Conclusion.
Therefore, the mass of \( CO_2 \) produced is approximately 11.44 gm.
Final Answer: 11.44