Question

1. balance the following reactions: (2 marks)

a) ko₂(s) + co₂(g) → k₂co₃(s) + o₂(g)

Explanation:

Step1: Balance potassium atoms

On the left - hand side, we have \(KO_{2}\) and on the right - hand side, we have \(K_{2}CO_{3}\). Let's start by balancing potassium. If we put a coefficient of \(4\) in front of \(KO_{2}\) and a coefficient of \(2\) in front of \(K_{2}CO_{3}\), the potassium atoms are balanced (\(4K\) on the left and \(2\times2 = 4K\) on the right). So we have \(4KO_{2}(s)+CO_{2}(g)
ightarrow 2K_{2}CO_{3}(s)+O_{2}(g)\)

Step2: Balance carbon atoms

After balancing potassium, we look at carbon. We have \(CO_{2}\) on the left and \(K_{2}CO_{3}\) on the right. If we put a coefficient of \(2\) in front of \(CO_{2}\), the carbon atoms are balanced (\(2C\) on the left and \(2C\) on the right). Now the equation is \(4KO_{2}(s)+2CO_{2}(g)
ightarrow 2K_{2}CO_{3}(s)+O_{2}(g)\)

Step3: Balance oxygen atoms

On the left - hand side, the number of oxygen atoms: from \(4KO_{2}\), we have \(4\times2=8\) oxygen atoms, and from \(2CO_{2}\), we have \(2\times2 = 4\) oxygen atoms. So in total \(8 + 4=12\) oxygen atoms on the left. On the right - hand side, from \(2K_{2}CO_{3}\), we have \(2\times3 = 6\) oxygen atoms. Let the coefficient of \(O_{2}\) be \(x\). Then the number of oxygen atoms on the right is \(6+2x\). Solving \(6 + 2x=12\) gives \(x = 3\). So the balanced equation is \(4KO_{2}(s)+2CO_{2}(g)
ightarrow 2K_{2}CO_{3}(s)+3O_{2}(g)\)

Answer:

\(4KO_{2}(s)+2CO_{2}(g)
ightarrow 2K_{2}CO_{3}(s)+3O_{2}(g)\)