Question

3 multiple choice 2 points balance this reaction: \\_\\_\\_k+\\_\\_\\_cl₂→\\_\\_\\_kcl 2k + 1cl₂→2kcl 2k + 2cl₂→2kcl 1k + 1cl₂→1kcl 1k + 2cl₂→2kcl

Explanation:

Step1: Count atoms on both sides

On the left - hand side (reactants): \(K\) atoms: \(x\) (from \(xK\)), \(Cl\) atoms: \(2y\) (from \(yCl_2\)). On the right - hand side (products): \(K\) atoms: \(z\) (from \(zKCl\)), \(Cl\) atoms: \(z\) (from \(zKCl\)).

Step2: Balance \(Cl\) atoms first

For \(Cl\) atoms: \(2y = z\). For \(K\) atoms: \(x=z\). Let's start by balancing \(Cl\). If we take \(y = 1\), then \(z = 2\) (to balance \(Cl\) atoms: \(Cl_2\) has 2 \(Cl\) atoms and \(2KCl\) has 2 \(Cl\) atoms).

Step3: Balance \(K\) atoms

Since \(z = 2\) (from balancing \(Cl\) atoms), then \(x = 2\) (to balance \(K\) atoms: \(2K\) has 2 \(K\) atoms and \(2KCl\) has 2 \(K\) atoms).

Answer:

\(2K + 1Cl_2
ightarrow2KCl\) (the first option)