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 = (mass)/(molar mass)`
∴
`mol of ZnCl_(2) = (2.00 g)/([(65)+(35.5 * 2)g/mol)`
=
`(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) <span>` = 0.0147 mol

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