Question

Balance the following chemical equations
(a) \[ \mathrm{C_6H_{12}O_6 + 6\,O_2 \;\longrightarrow\; 6\,CO_2 + 6\,H_2O} \]
\[ \mathrm{CaCO_3 + 2\,HCl \;\longrightarrow\; CaCl_2 + H_2O + CO_2} \]
\[ \mathrm{N_2 + 3\,H_2 \;\longrightarrow\; 2\,NH_3} \]
\[ \mathrm{MnO_2 + 4\,HCl \;\longrightarrow\; MnCl_2 + Cl_2 + 2\,H_2O} \]

Answer 1

(a) \[ \mathrm{C_6H_{12}O_6 + 6\,O_2 \;\longrightarrow\; 6\,CO_2 + 6\,H_2O} \]

Steps:

• Balance C: \(6\) on left → put \(6\) in front of \(CO_2\).
• Balance H: \(12\) on left → put \(6\) in front of \(H_2O\).
• Balance O: right has \(12+6=18\); left already has \(6\) in glucose → need \(12\) more → \(6O_2\).

(b) \[ \mathrm{CaCO_3 + 2\,HCl \;\longrightarrow\; CaCl_2 + H_2O + CO_2} \]

Steps:

• Ca and \(CO_3\) go to \(CaCl_2\), \(CO_2\) and \(H_2O\).
• Need two Cl and two H → coefficient \(2\) before \(HCl\).

(c) \[ \mathrm{N_2 + 3\,H_2 \;\longrightarrow\; 2\,NH_3} \]

Steps:

• N: put \(2\) on \(NH_3\) to match \(N_2\).
• H: right now \(6\) H → need \(3H_2\) on left.

(d) \[ \mathrm{MnO_2 + 4\,HCl \;\longrightarrow\; MnCl_2 + Cl_2 + 2\,H_2O} \]

Steps:

• Mn already balanced.
• O: two O on left → make \(2H_2O\) on right.
• Cl: total on right \(=2\) (in \(MnCl_2\)) \(+\) \(2\) (in \(Cl_2\)) \(=4\) → need \(4HCl\).
• H: now \(4\) H on left → matches \(2H_2O\) on right.