Warm Up 1/16 1. Balance the following equation: CH_(4)+O_(2)arrow CO_(2)+H_(2)O What is the coefficient in front of carbon tetrahydride? 2. What type of reaction is the formula above? 3. What type of reaction is the following formula? 2AlBr_(3)+3K_(2)SO_(4)- gt 6KBr+Al_(2)(SO_(4))_(3)

1. To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. In this case, we have 1 carbon atom, 4 hydrogen atoms, and 2 oxygen atoms on the left side, and 1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms on the right side. To balance the hydrogen atoms, we need to have 2 water molecules on the right side. This gives us the balanced equation: $CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O$. Therefore, the coefficient in front of carbon tetrahydride (CH4) is 1.<br /><br />2. The reaction $CH_{4}+2O_{2}\rightarrow CO_{2}+2H_{2}O$ is a combustion reaction. In a combustion reaction, a hydrocarbon (in this case, methane) reacts with oxygen to produce carbon dioxide and water.<br /><br />3. The reaction $2AlBr_{3}+3K_{2}SO_{4}\rightarrow 6KBr+Al_{2}(SO_{4})_{3}$ is a double displacement reaction. In a double displacement reaction, the cations and anions of two different compounds exchange places to form two new compounds.