which coefficient is needed in front of nano3...

# Explanation: ## Step1: Count atoms of each element on reactant side - For \(Na\): There are \(2\) \(Na\) atoms in \(Na_2S\). - For \(S\): There is \(1\) \(S\) atom in \(Na_2S\) and \(1\) \(S\) atom in \(ZnS\) (already balanced). - For \(Zn\): There is \(1\) \(Zn\) atom in \(Zn(NO_3)_2\) and \(1\) \(Zn\) atom in \(ZnS\) (already balanced). - For \(NO_3^-\): There are \(2\) \(NO_3^-\) groups in \(Zn(NO_3)_2\). ## Step2: Balance \(Na\) and \(NO_3^-\) on product side - Since there are \(2\) \(Na\) atoms (from \(Na_2S\)) and \(2\) \(NO_3^-\) groups (from \(Zn(NO_3)_2\)), we need \(2\) \(NaNO_3\) molecules. Because in \(NaNO_3\), each molecule has \(1\) \(Na\) atom and \(1\) \(NO_3^-\) group. So, \(2\times1 = 2\) for \(Na\) (matches \(Na_2S\)) and \(2\times1=2\) for \(NO_3^-\) (matches \(Zn(NO_3)_2\)). The balanced equation is \(Na_2S+Zn(NO_3)_2\rightarrow ZnS + 2NaNO_3\) # Answer: 2