Question

complete the atom inventory for the reaction below.
c₄h₁₀ + o₂ → co₂ + h₂o

Explanation:

Step1: Count oxygen atoms in reactants

In reactants, \(O_2\) has \(2\) oxygen atoms per molecule. But we need to consider the stoichiometry after balancing. For now, just note the formula.

Step2: Count oxygen atoms in products

In \(CO_2\), each molecule has \(2\) oxygen atoms. In \(H_2O\), each molecule has \(1\) oxygen atom.

Answer:

Reactants (O): \(2\) (from \(O_2\) formula, before balancing). Products (O): \(3\) ( \(2\) in \(CO_2\) and \(1\) in \(H_2O\) as per un - balanced equation). But for a balanced equation \(2C_4H_{10}+13O_2 = 8CO_2 + 10H_2O\), reactants (O): \(26\) (\(13\times2\)), products (O): \(26\) (\(8\times2 + 10\times1\)). Since the question seems to be about the un - balanced equation (as \(C\) and \(H\) counts are as per formula units), reactants (O) from \(O_2\) is \(2\) (per \(O_2\) molecule), products (O): \(3\) (\(2\) in \(CO_2\) and \(1\) in \(H_2O\) for one set of un - balanced products).