Complete and balance the equations for the following reactions. a. \(\mathrm{Cs}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow\) b. \(\mathrm{Na}(s)+\mathrm{Cl}_{2}(g) \rightarrow\)

In this reaction, cesium (Cs) reacts with water (H2O). Alkali metals reacting with water usually produce a metal hydroxide and hydrogen gas. In this case, the reaction produces cesium hydroxide, CsOH, and hydrogen gas, H2. Now we can write the unbalanced equation: \(Cs(s) + H_{2}O(l) \rightarrow CsOH(aq) + H_{2}(g)\) To balance the equation, we need to adjust the coefficients to ensure the number of atoms of each element is the same on both sides of the arrow. Here, we notice that the equation is already balanced: \(Cs(s) + H_{2}O(l) \rightarrow CsOH(aq) + H_{2}(g)\)

In this reaction, sodium (Na) reacts with chlorine gas (Cl2), which is a halogen reaction. Alkali metals reacting with halogens typically produce metal halides. In this case, the product is sodium chloride (NaCl). Now we can write the unbalanced equation: \(Na(s) + Cl_{2}(g) \rightarrow NaCl(s)\) To balance the equation, we need to adjust the coefficients to ensure the number of atoms of each element is the same on both sides of the arrow. We notice that there are 2 chlorine atoms on the left side, and only 1 on the right side. Therefore, we need to multiply the NaCl by 2 on the right side: \(Na(s) + Cl_{2}(g) \rightarrow 2NaCl(s)\) Now, we also need to multiply the Na on the left side by 2: \(2Na(s) + Cl_{2}(g) \rightarrow 2NaCl(s)\) The final balanced equation is: \(2Na(s) + Cl_{2}(g) \rightarrow 2NaCl(s)\)