Question

1. balance the following reactions: (2 marks)

2a) 2 ko₂(s) + 2 co₂(g) → 2 k₂co₃(s) + 3 o₂(g)
b) 2 c₈h₁₈(l) + 25 o₂(g) → 16 co₂(g) + 18 h₂o(g)

Explanation:

Step1: Balance the first reaction (a)

We use the method of balancing atoms. For the reaction \(KO_2(s)+CO_2(g)\to K_2CO_3(s)+O_2(g)\)

• Potassium (K) atoms:

On the left - hand side, we have \(1\) K in \(KO_2\). On the right - hand side, we have \(2\) K in \(K_2CO_3\). So we put a coefficient of \(4\) in front of \(KO_2\) and \(2\) in front of \(K_2CO_3\). The equation becomes \(4KO_2(s)+CO_2(g)\to 2K_2CO_3(s)+O_2(g)\)

• Carbon (C) atoms:

After putting the coefficient for K, we have \(1\) C in \(CO_2\) on the left and \(2\) C in \(2K_2CO_3\) on the right. So we put a coefficient of \(2\) in front of \(CO_2\). The equation is \(4KO_2(s)+2CO_2(g)\to 2K_2CO_3(s)+O_2(g)\)

• Oxygen (O) atoms:

On the left - hand side, \(4KO_2\) has \(4\times2 = 8\) O atoms and \(2CO_2\) has \(2\times2=4\) O atoms, a total of \(8 + 4=12\) O atoms. On the right - hand side, \(2K_2CO_3\) has \(2\times3 = 6\) O atoms. Let the coefficient of \(O_2\) be \(x\). Then \(6+2x=12\), so \(2x = 6\) and \(x = 3\). The balanced equation is \(4KO_2(s)+2CO_2(g)=2K_2CO_3(s)+3O_2(g)\)

Step2: Balance the second reaction (b)

For the reaction \(C_8H_{18}(l)+O_2(g)\to CO_2(g)+H_2O(g)\)

• Carbon (C) atoms:

We have \(8\) C in \(C_8H_{18}\). So we put a coefficient of \(8\) in front of \(CO_2\). The equation is \(C_8H_{18}(l)+O_2(g)\to 8CO_2(g)+H_2O(g)\)

• Hydrogen (H) atoms:

We have \(18\) H in \(C_8H_{18}\). So we put a coefficient of \(9\) in front of \(H_2O\). The equation is \(C_8H_{18}(l)+O_2(g)\to 8CO_2(g)+9H_2O(g)\)

• Oxygen (O) atoms:

On the right - hand side, \(8CO_2\) has \(8\times2 = 16\) O atoms and \(9H_2O\) has \(9\times1=9\) O atoms, a total of \(16 + 9=25\) O atoms. On the left - hand side, let the coefficient of \(O_2\) be \(x\). Then \(2x=25\), so \(x=\frac{25}{2}\). But we want whole - number coefficients. Multiply the entire equation by \(2\): \(2C_8H_{18}(l)+25O_2(g)=16CO_2(g)+18H_2O(g)\)

Answer:

a) \(4KO_2(s)+2CO_2(g)=2K_2CO_3(s)+3O_2(g)\)
b) \(2C_8H_{18}(l)+25O_2(g)=16CO_2(g)+18H_2O(g)\)