Question

cao + c→ cac₂ + co
total # before (reactants) elements total # after (products)
ca
c
o

Explanation:

Step1: Balance Ca

Let the coefficient of \(CaO\) be \(x\) and \(CaC_{2}\) be \(y\). For Ca: \(x = y\). Let \(x=y = 1\) for simplicity initially.

Step2: Balance O

From \(CaO\), if \(x = 1\), then number of O atoms in reactants is \(1\). In products, O is in \(CO\). Let the coefficient of \(CO\) be \(z\). So \(z=1\) (from O - balance).

Step3: Balance C

In reactants, C has coefficient \(a\). In products, C is in \(CaC_{2}\) (\(2\) atoms per formula unit, coefficient \(y = 1\)) and \(CO\) ( \(1\) atom per formula unit, coefficient \(z = 1\)). So total C in products is \(2\times1+1\times1=3\). So \(a = 3\).

Check the equation: \(CaO+3C
ightarrow CaC_{2}+CO\)

For Ca: \(1\) (reactants) \(=1\) (products)
For C: \(3\) (reactants) \(=2 + 1=3\) (products)
For O: \(1\) (reactants) \(=1\) (products)

Answer:

Total # Before (reactants)	Elements	Total # After (Products)
\(3\)	\(C\)	\(2 + 1=3\)
\(1\)	\(O\)	\(1\)

The balanced equation is \(CaO + 3C
ightarrow CaC_{2}+CO\)