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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?

Respuesta :

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

[tex]Moles of Na_{2}Zn(OH)_{4} :[/tex]
[tex]ratio of ZnCl_{2} : Na_{2}Zn(OH)_{4} [/tex]
                       1      :       1
∴ [tex]Moles of Na_{2}Zn(OH)_{4} [/tex] = 0.0147 mol

Mass = Molar Mass * Mol
∴ [tex]Mass of Na_{2}Zn(OH)_{4} [/tex] = [(23 * 2)+(65)+(16 * 4) + (1*4)] g/mol * 0.0147 mol
                                                       = 179 g/mol * 0.0147 mol
                                                       = 2.63 g

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