91Ó°ÊÓ

The given equation is \( \mathrm{SO}_{2}+\mathrm{O}_{2} \longrightarrow \mathrm{SO}_{3} \). Start by balancing the number of sulfur atoms. Both sides have one sulfur atom already.Next, balance the oxygen atoms. The right side has 3 oxygen atoms in \( \mathrm{SO}_{3} \), and the left has 2 in \( \mathrm{SO}_{2} \) and 2 in \( \mathrm{O}_2 \). Adjust the coefficients accordingly. The balanced equation is: \[ 2\mathrm{SO}_{2} + \mathrm{O}_{2} \longrightarrow 2\mathrm{SO}_{3} \]

The given equation is \( \mathrm{NH}_{3}+\mathrm{O}_{2} \longrightarrow\mathrm{N}_{2}+\mathrm{H}_{2} \mathrm{O} \). Start by balancing the nitrogen atoms. For nitrogen, there are 2 on the right side, so we need 2 \( \mathrm{NH}_{3} \) on the left. Next, balance the hydrogen atoms. The right side should have 6 hydrogens which are in 3 water molecules. This needs a coefficient of 3 for \( \mathrm{H}_2\mathrm{O} \).Finally, balance the oxygen atoms. The left should have 3 \( \mathrm{O}_2 \) to provide the necessary 3 extra oxygens.The balanced equation is: \[ 4 \mathrm{NH}_{3} + 3 \mathrm{O}_{2} \longrightarrow 2 \mathrm{N}_{2} + 6 \mathrm{H}_{2} \mathrm{O} \]

The given equation is \( \mathrm{C}_{4} \mathrm{H}_{10}+\mathrm{O}_{2} \longrightarrow\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O} \). Start by balancing the carbon atoms. There are 4 carbons in \( \mathrm{C}_{4}\mathrm{H}_{10} \), so place 4 in front of \( \mathrm{CO}_2 \).Next, balance the hydrogen atoms. There are 10 hydrogens in \( \mathrm{C}_{4}\mathrm{H}_{10} \), needing 5 \( \mathrm{H}_2\mathrm{O} \). Finally, balance the oxygen atoms. The right side now has 8 oxygen from \( \mathrm{CO}_2 \) and 5 from \( \mathrm{H}_2\mathrm{O} \), totaling 13. Therefore, use \( \frac{13}{2} \mathrm{O}_{2} \) or multiply the whole equation by 2 to get integers.The balanced equation is: \[ 2 \mathrm{C}_{4} \mathrm{H}_{10} + 13 \mathrm{O}_{2} \longrightarrow 8 \mathrm{CO}_{2} + 10 \mathrm{H}_{2} \mathrm{O} \]

The given equation is \( \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow\mathrm{PbO}+\mathrm{NO}_{2}+\mathrm{O}_{2} \). Start by balancing the lead atoms. There is one \( \mathrm{Pb} \) on both sides.Next, balance the nitrogen atoms. The left side has 2 nitrogen (inside \( \mathrm{NO}_3 \)), so use 2 \( \mathrm{NO}_{2} \) on the right side.Balance the oxygen atoms. The left has 6 oxygen atoms, and on the right with 1 \( \mathrm{O}_2 \) and 2 \( \mathrm{NO}_{2} \) sum up to 6.The balanced equation is: \[ \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow \mathrm{PbO} + 2 \mathrm{NO}_{2} + \frac{1}{2} \mathrm{O}_{2} \] or multiplying through by 2: \[ 2 \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2} \rightarrow 2 \mathrm{PbO} + 4 \mathrm{NO}_{2} + \mathrm{O}_{2} \]

The given equation is \( \mathrm{Al}+\mathrm{Fe}_{3} \mathrm{O}_{4} \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{Fe} \). Start by balancing aluminum by placing a 2 in front of \( \mathrm{Al} \) and \( \mathrm{Al}_2\mathrm{O}_3 \).Next, balance the iron atoms. The right must match the 3 iron atoms in \( \mathrm{Fe}_3\mathrm{O}_4 \).Then, verify the oxygen atoms for balance. Both sides correspond with \( \mathrm{Fe}_{3} \mathrm{O}_{4} \).The balanced equation is: \[ 8\mathrm{Al} + 3\mathrm{Fe}_{3} \mathrm{O}_{4} \longrightarrow 4\mathrm{Al}_{2} \mathrm{O}_{3} + 9\mathrm{Fe} \]

The given equation is \( \mathrm{Sr} + \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \). Start by balancing strontium, which is already balanced.Next, balance hydrogen. The use of 2 \( \mathrm{H}_{2} \mathrm{O} \) provides 4 hydrogen to match \( \mathrm{Sr}(\mathrm{OH})_{2} \) and \( \mathrm{H}_{2} \).Finally, balance oxygen by utilizing it from \( \mathrm{H}_{2} \mathrm{O} \).The balanced equation is: \[ \mathrm{Sr} + 2\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Sr}(\mathrm{OH})_{2} + \mathrm{H}_{2} \]