\(4NaOH _{(aq)} + ZnCl_{2}_{(aq)} ----\ \textgreater \ Na_{2}Zn(OH)_{4}_{(ppt)} + 2NaCl (aq)\)
\(mol = \frac{mass}{molar mass}\)
∴ \(mol of ZnCl_{2} = \frac{2.00 g}{[(65)+(35.5 * 2)g/mol}\)
= \(\frac{2.00 g}{136 g/mol}\)
= 0.0147 mol
\(Moles of Na_{2}Zn(OH)_{4} :\)
\(ratio of ZnCl_{2} : Na_{2}Zn(OH)_{4}\)
1 : 1
∴ \(Moles of Na_{2}Zn(OH)_{4} \) = 0.0147 mol
Mass = Molar Mass Mol
∴ \(Mass of Na_{2}Zn(OH)_{4} \) = [(23 2)+(65)+(16 4) + (14)] g/mol 0.0147 mol
= 179 g/mol 0.0147 mol
= 2.63 g
