Question

select the correct answer. which equation obeys the law of conservation of mass? a. 2c₄h₁₀ + 2cl₂ + 12o₂ → 7co₂ + ccl₄ + 10h₂o b. c₄h₁₀+ cl₂ + 2o₂ → 7co₂ + ccl₄ + 8h₂o c. 2c₄h₁₀+ 2cl₂ + 12o₂ → 4co₂ + ccl₄ + h₂o d. 2c₄h₁₀+ 2cl₂ + 6o₂ → co₂ + ccl₄ + 6h₂o e. 4c₄h₁₀+ 4cl₂ + 14o₂ → 14co₂ + ccl₄ + 20h₂o

Explanation:

Step1: Check carbon atoms

For option A:
On the left - hand side, in \(2C_4H_{10}\), the number of carbon atoms is \(2\times4 = 8\). On the right - hand side, in \(7CO_2\), the number of carbon atoms is 7. This option is incorrect.
For option B:
In \(C_4H_{10}\), there are 4 carbon atoms on the left - hand side. In \(7CO_2\), there are 7 carbon atoms on the right - hand side. This option is incorrect.
For option C:
In \(2C_4H_{10}\), there are \(2\times4=8\) carbon atoms on the left - hand side. In \(4CO_2\), there are 4 carbon atoms on the right - hand side. This option is incorrect.
For option D:
In \(2C_4H_{10}\), there are \(2\times4 = 8\) carbon atoms on the left - hand side. In \(CO_2\), there is 1 carbon atom on the right - hand side. This option is incorrect.
For option E:
In \(4C_4H_{10}\), the number of carbon atoms on the left - hand side is \(4\times4=16\). In \(14CO_2\), the number of carbon atoms on the right - hand side is 14. This option is incorrect.

Let's check all elements systematically for option A:
Carbon (C):
Left - hand side: In \(2C_4H_{10}\), \(n_C=2\times4 = 8\). Right - hand side: In \(7CO_2\), \(n_C = 7\).
Hydrogen (H):
Left - hand side: In \(2C_4H_{10}\), \(n_H=2\times10 = 20\). Right - hand side: In \(10H_2O\), \(n_H=10\times2=20\).
Chlorine (Cl):
Left - hand side: In \(2Cl_2\), \(n_{Cl}=2\times2 = 4\). Right - hand side: In \(CCl_4\), \(n_{Cl}=4\).
Oxygen (O):
Left - hand side: In \(12O_2\), \(n_O=12\times2 = 24\). Right - hand side: In \(7CO_2\), \(n_O = 7\times2=14\), in \(10H_2O\), \(n_O=10\times1 = 10\), total \(n_O=14 + 10=24\).

Answer:

A. \(2C_4H_{10}+2Cl_2 + 12O_2
ightarrow7CO_2+CCl_4 + 10H_2O\)