Question

write a balanced chemical equation for the reaction of zncl2 with excess NaOH to produce Na2Zn(oh)4 sodium zincate. what mass of sodium zincate can be produced from 2.00 g of ZNCl2 with excess Naoh by this reaction?

Answer 1

$4NaOH _{(aq)} + ZnCl_{2}_{(aq)} ----\ \textgreater \ Na_{2}Zn(OH)_{4}_{(ppt)} + 2NaCl (aq)$
 
$mol = \frac{mass}{molar mass}$
∴  $mol of ZnCl_{2} = \frac{2.00 g}{[(65)+(35.5 * 2)g/mol}$
                                    =  $\frac{2.00 g}{136 g/mol}$
                                    =  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) + (1*4)] g/mol * 0.0147 mol
                                                       = 179 g/mol * 0.0147 mol
                                                       = 2.63 g