Question

1. c₃h₈ + o₂ → co₂ + h₂o
2. ai(oh)₃ + hbr → aibr₃ + h₂o

Explanation:

Step1: Balance the carbon atoms

For the reaction \(C_{3}H_{8}+O_{2}\to CO_{2}+H_{2}O\), there are 3 carbon atoms in \(C_{3}H_{8}\). So, we put a 3 in front of \(CO_{2}\). The equation becomes \(C_{3}H_{8}+O_{2}\to3CO_{2}+H_{2}O\).

Step2: Balance the hydrogen atoms

There are 8 hydrogen atoms in \(C_{3}H_{8}\). So, we put a 4 in front of \(H_{2}O\). The equation becomes \(C_{3}H_{8}+O_{2}\to3CO_{2}+4H_{2}O\).

Step3: Balance the oxygen atoms

On the right - hand side, there are \(3\times2 + 4\times1=6 + 4 = 10\) oxygen atoms. So, we put a 5 in front of \(O_{2}\). The balanced equation is \(C_{3}H_{8}+5O_{2}=3CO_{2}+4H_{2}O\).

For the reaction \(Al(OH)_{3}+HBr\to AlBr_{3}+H_{2}O\)

Step1: Balance the bromine atoms

There are 3 bromine atoms in \(AlBr_{3}\). So, we put a 3 in front of \(HBr\). The equation becomes \(Al(OH)_{3}+3HBr\to AlBr_{3}+H_{2}O\).

Step2: Balance the hydrogen and oxygen atoms

On the left - hand side, there are \(3 + 3=6\) hydrogen atoms and 3 oxygen atoms from \(Al(OH)_{3}\). On the right - hand side, in \(AlBr_{3}\) there are no \(O\) or \(H\) (except in \(H_{2}O\)). From the hydrogen atoms: if we put a 3 in front of \(H_{2}O\), we have \(3\times2 = 6\) hydrogen atoms and \(3\times1=3\) oxygen atoms. The balanced equation is \(Al(OH)_{3}+3HBr = AlBr_{3}+3H_{2}O\).

Answer:

1. \(1C_{3}H_{8}+5O_{2}=3CO_{2}+4H_{2}O\)
2. \(1Al(OH)_{3}+3HBr = 1AlBr_{3}+3H_{2}O\)