Question

1. balance the following chemical equations: (2 marks each)

c₃h₈(g) + o₂(g) → co₂(g) + h₂o(g)
bacl₂(aq) + k₃po₄(aq) → kcl(aq) + ba₃(po₄)₂(s)

Explanation:

Step1: Balance the first equation

We start with the equation $C_3H_8(g)+O_2(g)
ightarrow CO_2(g)+H_2O(g)$. First, balance the carbon atoms. There are 3 carbon atoms in $C_3H_8$, so we put a 3 in front of $CO_2$: $C_3H_8(g)+O_2(g)
ightarrow 3CO_2(g)+H_2O(g)$. Then balance the hydrogen atoms. There are 8 hydrogen atoms in $C_3H_8$, so we put a 4 in front of $H_2O$: $C_3H_8(g)+O_2(g)
ightarrow 3CO_2(g)+4H_2O(g)$. Now, balance the oxygen atoms. On the right - hand side, we have $3\times2 + 4\times1=10$ oxygen atoms. So we put a 5 in front of $O_2$ on the left - hand side. The balanced equation is $C_3H_8(g)+5O_2(g)
ightarrow 3CO_2(g)+4H_2O(g)$.

Step2: Balance the second equation

For the equation $BaCl_2(aq)+K_3PO_4(aq)
ightarrow KCl(aq)+Ba_3(PO_4)_2(s)$. First, balance the barium atoms. There are 3 barium atoms in $Ba_3(PO_4)_2$, so we put a 3 in front of $BaCl_2$: $3BaCl_2(aq)+K_3PO_4(aq)
ightarrow KCl(aq)+Ba_3(PO_4)_2(s)$. Then balance the phosphate groups. There are 2 phosphate groups in $Ba_3(PO_4)_2$, so we put a 2 in front of $K_3PO_4$: $3BaCl_2(aq)+2K_3PO_4(aq)
ightarrow KCl(aq)+Ba_3(PO_4)_2(s)$. Now, balance the potassium and chlorine atoms. There are 6 potassium atoms and 6 chlorine atoms on the left - hand side, so we put a 6 in front of $KCl$. The balanced equation is $3BaCl_2(aq)+2K_3PO_4(aq)
ightarrow 6KCl(aq)+Ba_3(PO_4)_2(s)$.

Answer:

1. $C_3H_8(g)+5O_2(g)

ightarrow 3CO_2(g)+4H_2O(g)$

1. $3BaCl_2(aq)+2K_3PO_4(aq)

ightarrow 6KCl(aq)+Ba_3(PO_4)_2(s)$